Eigenvalues and Eigenvectors

We are looking for a vector \(\displaystyle u=\left[\begin{array}{c}x\\y\end{array}\right]\) and a scalar \(\lambda\in\mathbb R\) such that \(Au=\lambda u\). The last equation is equivalent to the system: \begin{eqnarray*} 5x+4y&=&\lambda x\\ -4x-5y&=&\lambda y. \end{eqnarray*} The previous system is equivalent to: \begin{eqnarray*} (5-\lambda)x+4y&=&0\\ -4x+(-5-\lambda)y&=&0. \end{eqnarray*} If we multiply the first equation by \(5+\lambda\), the second equation by \(4\) and add the obtained two equations we deduce: \((25-\lambda^2-16)x=0\). If \(9-\lambda^2\neq 0\), then we must have \(x=0\). This would imply that \(y=0\) which would lead to a trivial eigenvector. Assume that \(9-\lambda^2=0\). Then we have \(\lambda\in\{-3,3\}\), and these are the two eigenvalues.

The eigenvalue \(\lambda_1=3\) corresponds to the eigenvector whose coordinates \((x,y)\) satisfy the system: \begin{eqnarray*} 2x+4y&=&0\\ -4x-8y&=&0. \end{eqnarray*} A non-trivial solution \((-2,1)\) corresponds to the eigenvector \(\displaystyle u_1=\left[\begin{array}{c}-2\\1\end{array}\right]\).

Similarly, the eigenvalue \(\lambda_2=-3\) gives us the system: \begin{eqnarray*} 8x+4y&=&0\\ -4x-2y&=&0. \end{eqnarray*} This gives us another eigenvector \(\displaystyle u_2=\left[\begin{array}{c}-1\\2\end{array}\right]\).

The operator \(A^n\) is uniquely determined by the vectors \(A^nv\) and \(A^nw\) for any two vectors \(v,w\in\mathbb R^2\) that form a basis for \(\mathbb R^2\).

Assume that there are two eigenvectors of \(A\) that form a basis for \(\mathbb R^2\). Assume that \(v_1,v_2\in\mathbb R^2\) and \(\lambda_1,\lambda_2\in\mathbb R\) satisfy \(Av_1=\lambda_1v_1\) and \(Av_2=\lambda_2v_2\). Then we have \[A^nv_1=A^{n-1}(Av_1)=A^{n-1}(\lambda_1v_1)=\lambda_1A^{n-1}v_1=\lambda_1A^{n-2}(Av_1))=\lambda_1^2A^{n-2}v_1=\cdots=\lambda_1^nv_1,\] and similarly \(A^nv_2=\lambda_2^nv_2\).

From Example 1 we know that \(\displaystyle u_1=\left[\begin{array}{c}-2\\1\end{array}\right]\) and \(\displaystyle u_2=\left[\begin{array}{c}-1\\2\end{array}\right]\) are the eigenvectors of \(A\) and that their corresponding eigenvalues are \(3\) and \(-3\). Therefore \(\displaystyle A^n\left[\begin{array}{c}-2\\1\end{array}\right]=\left[\begin{array}{c}-2\cdot 3^n\\3^n\end{array}\right]\) and \(\displaystyle A^n\left[\begin{array}{c}-1\\2\end{array}\right]=\left[\begin{array}{c}-1\cdot (-3)^n\\2\cdot (-3)^n\end{array}\right]\).

It remains to find the matrix of \(A^n\). In order to do that we need to find \(\displaystyle A^n\left[\begin{array}{c}1\\0\end{array}\right]\) and \(\displaystyle A^n\left[\begin{array}{c}0\\1\end{array}\right]\).

• Calculating \(\displaystyle A^n\left[\begin{array}{c}1\\0\end{array}\right]\). We need to find scalars \(\alpha_1\) and \(\alpha_2\) such that \(\displaystyle \left[\begin{array}{c}1\\0\end{array}\right]= \alpha_1u_1+\alpha_2u_2\). This gives us the system of equations: \begin{eqnarray*} -2\alpha_1-\alpha_2&=&1\\ \alpha_1+2\alpha_2&=&0. \end{eqnarray*} Multiplying the second equation by 2 and adding it to the first implies \(\alpha_2=\frac13\). Substituting this value into the second equation yields \(\alpha_1=-\frac23\). Thus \[A^n\left[\begin{array}{c}1\\0\end{array}\right]=-\frac23A^nu_1+\frac13A^nu_2= \left[\begin{array}{c} \frac23\cdot 2\cdot 3^n+\frac13\cdot (-(-3)^n)\\ -\frac23\cdot 3^n+\frac13\cdot 2\cdot (-3)^n \end{array}\right] =\left[\begin{array}{c} 4\cdot 3^{n-1}+(-3)^{n-1}\\ -2\cdot 3^{n-1}-2\cdot (-3)^{n-1} \end{array}\right] .\]

• Calculating \(\displaystyle A^n\left[\begin{array}{c}0\\1\end{array}\right]\). We need to find scalars \(\beta_1\) and \(\beta_2\) for which \(\displaystyle \left[\begin{array}{c}0\\1\end{array}\right]=\beta_1u_1+\beta_2 u_2\). The last equation is equivalent to the following system: \begin{eqnarray*} -2\beta_1-\beta_2&=&0\\ \beta_1+2\beta_2&=&1. \end{eqnarray*} If we multiply the second equation by \(2\) and add it to the first we obtain \(\beta_2=\frac23\). Substituting this value in the first equation yields \(\beta_1=-\frac13\). Therefore \[A^n\left[\begin{array}{c}0\\1\end{array}\right]=-\frac13A^nu_1+\frac23A^nu_2= \left[\begin{array}{c} \frac13\cdot 2\cdot 3^n+\frac23\cdot (-(-3)^n)\\ -\frac13\cdot 3^n+\frac43\cdot (-3)^n \end{array}\right] =\left[\begin{array}{c} 2\cdot 3^{n-1}+2\cdot (-3)^{n-1}\\ - 3^{n-1}-4\cdot (-3)^{n-1} \end{array}\right] .\]

Thus \[A^n=\left[\begin{array}{cc} 4\cdot 3^{n-1}+(-3)^{n-1} & 2\cdot 3^{n-1}+2\cdot (-3)^{n-1}\\ -2\cdot 3^{n-1}-2\cdot (-3)^{n-1} & - 3^{n-1}-4\cdot (-3)^{n-1} \end{array}\right], \] or, equivalently: \begin{eqnarray*} a_n&=&4\cdot 3^{n-1}+(-3)^{n-1} \\ b_n&=&2\cdot 3^{n-1}+2\cdot (-3)^{n-1}\\ c_n&=& -2\cdot 3^{n-1}-2\cdot (-3)^{n-1} \\ d_n&=& - 3^{n-1}-4\cdot (-3)^{n-1}. \end{eqnarray*}

Assume that \(\displaystyle v_i=\left[\begin{array}{c} v_{1i}\\v_{2i}\\\vdots\\v_{ni}\end{array}\right]\), and let \(\displaystyle P= \left[\begin{array}{ccccc} v_{11}&v_{12}&v_{13}&\cdots&v_{1n}\\ v_{21}&v_{22}&v_{23}&\cdots&v_{2n}\\ & & &\vdots& \\ v_{n1}&v_{n2}&v_{n3}&\cdots&v_{nn} \end{array}\right]\).

Let \(e_1\), \(\dots\), \(e_n\) be the standard basis of \(\mathbb R^n\). In other words, the vector \(e_i\) has \(1\) at the \(i\)-th position and \(0\) at every other position. Then we have \(Pe_i=v_i\) for each \(i\in\{1,2,\dots, n\}\). Thus \(APe_i=Av_i=\lambda_iv_i\) and \(P^{-1}APe_i=P^{-1}(\lambda_iv_i)=\lambda_iP^{-1}v_i=\lambda_ie_i\). Thus \[ P^{-1}AP(e_i)=\left[\begin{array}{c}0\\\vdots\\0\\\lambda_i\\0\\\vdots\\0\end{array}\right].\] The \(i\)-th row of the matrix of the operator \(P^{-1}AP\) must be equal to \(P^{-1}AP(e_i)\), hence the matrix of the operator \(P^{-1}AP\) must be \(\displaystyle \left[\begin{array}{ccccc} \lambda_1&0&0&\cdots&0\\0&\lambda_2&0&\cdots&0\\ 0&0&\lambda_3&\cdots&0\\&&&\vdots&\\ 0&0&0&\dots&\lambda_n\end{array}\right]\).

We will prove this theorem under the assumption that \(A\) has \(n\) linearly independent eigenvectors. The proof of the general case follows the same idea but requires the theory of Jordan canonical forms. In Example 5 at the end of this article we will see some techniques involving Jordan canonical forms.

According to Theorem 2 there is a matrix \(P\) such that \(P^{-1}AP=D\), where \(D\) is a diagonal matrix whose diagonal entries are the eigenvalues of \(A\). Then we have \(A=PDP^{-1}\) and \(A^k=PD^kP^{-1}\) for each \(k\in\mathbb N\). Thus \(\varphi_A(A)=P\varphi_A(D)P^{-1}\). However, \(\varphi_A(D)=0\), since the entries of \(D\) are the eigenvalues of \(A\), and each of them is a zero of the characteristic polynomial according to Theorem 1. This completes the proof.

Let \(\displaystyle A=\left[\begin{array}{cc} 5&4\\-4&-5\end{array}\right]\). Then we have for \(n\geq 0\): \(\displaystyle \left[\begin{array}{c}x_{n+1}\\y_{n+1}\end{array}\right]=A\left[\begin{array}{c}x_{n}\\y_{n}\end{array}\right]\). Therefore: \[\left[\begin{array}{c}x_{n}\\y_{n}\end{array}\right]=A\left[\begin{array}{c}x_{n-1}\\y_{n-1}\end{array}\right]= A^2\left[\begin{array}{c}x_{n-2}\\y_{n-2}\end{array}\right]=\cdots=A^n\left[\begin{array}{c}x_{0}\\y_{0}\end{array}\right]. \] In Example 2 we showed that \(\displaystyle A^n=\left[\begin{array}{cc} 4\cdot 3^{n-1}+(-3)^{n-1} & 2\cdot 3^{n-1}+2\cdot (-3)^{n-1}\\ -2\cdot 3^{n-1}-2\cdot (-3)^{n-1} & - 3^{n-1}-4\cdot (-3)^{n-1} \end{array}\right]\), hence: \[\left[\begin{array}{c}x_{n}\\y_{n}\end{array}\right]=\left[\begin{array}{cc} 4\cdot 3^{n-1}+(-3)^{n-1} & 2\cdot 3^{n-1}+2\cdot (-3)^{n-1}\\ -2\cdot 3^{n-1}-2\cdot (-3)^{n-1} & - 3^{n-1}-4\cdot (-3)^{n-1} \end{array}\right]\cdot \left[\begin{array}{c}3\\2\end{array}\right]= \left[\begin{array}{c} 4\cdot 3^{n}-(-3)^{n}+4\cdot 3^{n-1}+4\cdot (-3)^{n-1}\\ -2\cdot 3^{n}+2\cdot (-3)^{n} -2\cdot 3^{n-1}-8\cdot (-3)^{n-1} \end{array}\right]. \] Thus: \begin{eqnarray*} x_n&=&4\cdot 3^{n}-(-3)^{n}+4\cdot 3^{n-1}+4\cdot (-3)^{n-1}=16\cdot 3^{n-1}+7\cdot (-3)^{n-1}\\ y_n&=&-2\cdot 3^{n}+2\cdot (-3)^{n} -2\cdot 3^{n-1}-8\cdot (-3)^{n-1}=-8\cdot 3^{n-1}-14\cdot (-3)^{n-1}. \end{eqnarray*}

Let us denote \(G_n=F_{n+1}\). Then we have \(G_0=1\), \(F_0=0\) and for \(n\geq 0\) the following holds: \begin{eqnarray*} \begin{array}{ccccc} F_{n+1}&=& & &G_n\\ G_{n+1}&=&F_n&+&G_n, \end{array} \end{eqnarray*} or, equivalently \(\displaystyle \left[\begin{array}{c}F_{n+1}\\G_{n+1}\end{array}\right]= \left[\begin{array}{cc}0&1\\1&1\end{array}\right] \left[\begin{array}{c}F_{n}\\G_{n}\end{array}\right]\). Let us denote \(\displaystyle A=\left[\begin{array}{cc}0&1\\1&1\end{array}\right]\). Similarly to the argument presented in Example 3 we get \(\displaystyle \left[\begin{array}{c}F_{n}\\G_{n}\end{array}\right]=A^n\left[\begin{array}{c}F_{0}\\G_{0}\end{array}\right]\).

Therefore, our goal is to find \(A^n\). In order to do so, we will first find eigenvalues and eigenvectors of the matrix \(A\). The eigenvalues are the solutions of the equation \(\varphi_A(x)=0\), where \(\displaystyle \varphi_A(x)=\mbox{det }(A-xI)=-x(1-x)-1=x^2-x-1\). Since \(\displaystyle \varphi_A(x)=\left(x-\frac{1-\sqrt 5}2\right)\cdot \left(x-\frac{1+\sqrt 5}2\right)\), we get that the eigenvalues are \(\displaystyle \lambda_1=\frac{1-\sqrt 5}2\) and \(\displaystyle \lambda_2=\frac{1+\sqrt 5}2\).

We will now find an eigenvector corresponding to the eigenvalue \(\lambda_1\). Let us denote by \(\displaystyle u=\left[\begin{array}{c}u_1\\u_2\end{array}\right]\) the required eigenvector. Then we have \(-\lambda_1u_1+u_2=0\) and \(u_1+(1-\lambda_1)u_2=0\) and we may take \(\displaystyle u= \left[\begin{array}{c}1\\\lambda_1\end{array}\right]=\left[\begin{array}{c}1\\\frac{1-\sqrt 5}2\end{array}\right]\). Similarly, we find \(\displaystyle v= \left[\begin{array}{c}1\\\lambda_2\end{array}\right]=\left[\begin{array}{c}1\\\frac{1+\sqrt 5}2\end{array}\right]\). We have \(A^nu=\lambda_1^nu\) and \(A^nv=\lambda_2^nv\). In order to determine the matrix \(A^n\) it remains to find the vectors \(A^ne_1\) and \(A^ne_2\), where \(\displaystyle e_1=\left[\begin{array}{c}1\\0\end{array}\right]\) and \(\displaystyle e_2=\left[\begin{array}{c}0\\1\end{array}\right]\) are the elements of the standard basis of \(\mathbb R^2\). We will first express \(e_1\) in terms of \(u\) and \(v\). In order to do so we need to find scalars \(\alpha_1\) and \(\alpha_2\) such that \(e_1=\alpha_1u+\alpha_2v\). This gives us the system: \begin{eqnarray*} 1&=&\alpha_1+\alpha_2\\ 0&=&\alpha_1\cdot\frac{1-\sqrt 5}2+\alpha_2\cdot\frac{1+\sqrt 5}2. \end{eqnarray*} Solving the system gives us \(\displaystyle (\alpha_1,\alpha_2)=\left( \frac{1+\sqrt 5}{2\sqrt 5}, \frac{\sqrt 5-1}{2\sqrt 5} \right)\). We now obtain: \begin{eqnarray*} A^ne_1&=& A^n\left(\alpha_1u+\alpha_2 v\right)=\frac{1+\sqrt 5}{2\sqrt 5}\lambda_1^nu+\frac{\sqrt 5-1}{2\sqrt 5}\lambda_2^nv\\ &=&\frac1{\sqrt 5}\left[\begin{array}{c}\frac{(1+\sqrt 5)\lambda_1^n+(\sqrt 5-1)\lambda_2^n}2\\ \frac{(1+\sqrt 5)\lambda_1^{n+1}+(\sqrt 5-1)\lambda_2^{n+1}}2 \end{array}\right]. \end{eqnarray*} In order to find \(A^ne_2\) we first need to find the scalars \(\beta_1\) and \(\beta_2\) such that \(e_2=\beta_1u+\beta_2v\). The last equation becomes: \begin{eqnarray*} 0&=&\beta_1+\beta_2\\ 1&=&\beta_1\cdot\frac{1-\sqrt 5}2+\beta_2\cdot\frac{1+\sqrt 5}2. \end{eqnarray*} After solving the system we obtain \(\displaystyle (\beta_1,\beta_2)=\left(-\frac1{\sqrt 5},\frac1{\sqrt 5}\right)\). Then we have \begin{eqnarray*} A^ne_2&=& A^n\left(\beta_1u+\beta_2 v\right)=-\frac1{\sqrt 5}\lambda_1^nu+\frac1{\sqrt 5}\lambda_2^nv\\ &=&\frac1{\sqrt 5}\left[\begin{array}{c}\lambda_2^n-\lambda_1^n\\ \lambda_2^{n+2}-\lambda_1^{n+1} \end{array}\right]. \end{eqnarray*} Therefore \begin{eqnarray*} A^n&=& \frac1{\sqrt 5}\left[\begin{array}{cc}\frac{(1+\sqrt 5)\lambda_1^n+(\sqrt 5-1)\lambda_2^n}2&\lambda_2^n-\lambda_1^n\\ \frac{(1+\sqrt 5)\lambda_1^{n+1}+(\sqrt 5-1)\lambda_2^{n+1}}2&\lambda_2^{n+2}-\lambda_1^{n+1} \end{array}\right], \quad\quad \mbox{and}\quad\quad \left[\begin{array}{c}F_n\\G_n\end{array}\right]=A^n\left[\begin{array}{c}0\\1\end{array}\right]= \frac1{\sqrt 5}\left[\begin{array}{c}\lambda_2^n-\lambda_1^n\\\lambda_2^{n+1}-\lambda_1^{n+1}\end{array}\right]. \end{eqnarray*} Thus \[F_n=\frac1{\sqrt 5}\left(\frac{1+\sqrt 5}2\right)^n-\frac1{\sqrt 5}\left(\frac{1-\sqrt 5}2\right)^n.\]

• (a) The characteristic polynomial of \(A\) is \(\varphi_A(x)=(4-x)(2-x)+1=x^2-6x+9=(x-3)^2\), hence \(\lambda=3\) is the only eigenvalue of \(A\).

• (b) We need to find \(\displaystyle u=\left[\begin{array}{c} u_1\\u_2\end{array}\right]\) such that \(Au=3u\). This gives us the following system of equations: \begin{eqnarray*} u_1+u_2&=&0\\ -u_1-u_2&=&0, \end{eqnarray*} and each eigenvector of \(A\) must satisfy \(\displaystyle u=\left[\begin{array}{c} t\\-t\end{array}\right]\) for \(t\in\mathbb R\). We can take \(\displaystyle u=\left[\begin{array}{c} 1\\-1\end{array}\right]\).

• (c) Such \(w\) does not exist because each eigenvector \(w\) of \(A\) is of the form \(\displaystyle w=t\left[\begin{array}{c} 1\\-1\end{array}\right]=tu\).

• (d) Let us denote \(\displaystyle v=\left[\begin{array}{c} v_1\\v_2\end{array}\right]\). We will find \(v_1\) and \(v_2\). The numbers \(v_1\) and \(v_2\) must satisfy the following system of equations: \begin{eqnarray*} v_1+v_2&=&1\\ -v_1-v_u&=&-1. \end{eqnarray*} We can take \(\displaystyle v=\left[\begin{array}{c} 1\\0\end{array}\right]\).

• (e) Since the vectors \(u\) and \(v\) are linearly independent it suffices to find \(A^nu\) and \(A^nv\). Clearly, \(A^nu=\lambda^nu=3^nu\), and \begin{eqnarray*} A^nv&=& A^{n-1}(Av)=A^{n-1}\left(\lambda v+u\right)=\lambda A^{n-1}v + A^{n-1}u=\lambda A^{n-1}v+\lambda^{n-1}u\\ &=& \lambda A^{n-2}\left(Av\right)+\lambda^{n-1}u=\lambda A^{n-2}\left(\lambda v+u\right)+\lambda^{n-1}u=\lambda^2 A^{n-2}v +2\lambda^{n-1}u\\ &\vdots&\\ &=&\lambda^nv+n\lambda^{n-1}u=\left[\begin{array}{c}3^n+n3^{n-1}\\ -n3^{n-1}\end{array}\right]. \end{eqnarray*} We now need to find \(\displaystyle A^n\left[\begin{array}{c} 1\\0\end{array}\right]\) and \(\displaystyle A^n\left[\begin{array}{c} 0\\1\end{array}\right]\). We have already found that \(\displaystyle A^n\left[\begin{array}{c} 1\\0\end{array}\right]=\left[\begin{array}{c}3^n+n3^{n-1}\\ -n3^{n-1}\end{array}\right]\). In order to find the other vector we first express \(\displaystyle \left[\begin{array}{c} 0\\1\end{array}\right]\) in terms of \(u\) and \(v\): \(\displaystyle \left[\begin{array}{c} 0\\1\end{array}\right]=v-u\), hence: \begin{eqnarray*} A^n\left[\begin{array}{c} 0\\1\end{array}\right]&=&A^nv-A^nu=\left[\begin{array}{c}3^n+n3^{n-1}\\ -n3^{n-1}\end{array}\right]- \left[\begin{array}{c}3^n\\ -3^n\end{array}\right]=\left[\begin{array}{c}n3^{n-1}\\ 3^n-n3^{n-1}\end{array}\right]. \end{eqnarray*} Thus: \[A^n=\left[\begin{array}{cc} 3^n+n3^{n-1} & n3^{n-1}\\ -n3^{n-1} & 3^n-n3^{n-1}\end{array}\right]. \]

• (f) We solve the recursive system of equations by first expressing it in the form \(\displaystyle \left[\begin{array}{c}x_{n+1}\\y_{n+1}\end{array}\right]=A\left[\begin{array}{c}x_{n}\\y_{n}\end{array}\right]\). In the same way as in Example 3 we conclude that \(\displaystyle \left[\begin{array}{c}x_{n}\\y_{n}\end{array}\right]= A^n\left[\begin{array}{c}x_{0}\\y_{0}\end{array}\right]\), thus: \[\left[\begin{array}{c}x_{n}\\y_{n}\end{array}\right]= \left[\begin{array}{cc} 3^n+n3^{n-1} & n3^{n-1}\\ -n3^{n-1} & 3^n-n3^{n-1}\end{array}\right]\cdot \left[\begin{array}{c}2\\5\end{array}\right]=\left[\begin{array}{c}2\cdot 3^n+7n\cdot 3^{n-1}\\5\cdot 3^n-7n\cdot 3^{n-1}\end{array}\right],\] which is equivalent to: \begin{eqnarray*} x_n&=&2\cdot 3^n+7n\cdot 3^{n-1}\\ y_n&=&5\cdot 3^n-7n\cdot 3^{n-1}. \end{eqnarray*}

Since \begin{eqnarray*} Aw&=&\left[\begin{array}{ccc} 6 & 5 \\ 30 & 11 \end{array}\right]\left[\begin{array}{c}1\\-2\end{array}\right]=-4\left[\begin{array}{c}1\\-2\end{array}\right], \end{eqnarray*} the eigenvalue corresponding to the vector \(\left[\begin{array}{c}1\\-2\end{array}\right]\) is equal to \(-4\).

Notice that \(\displaystyle A=\left[\begin{array}{ccc} 9 & 5 \\ -8 & 6 \end{array}\right]\). Thus \(a_{11}+a_{22}=15\).

We want to find all vectors \(w=\left[\begin{array}{c} x\newline y \end{array}\right]\) such that \(Aw=\lambda w\) for some real number \(\lambda\).

The given equation is equivalent to the system: \begin{eqnarray*} \begin{array}{rrl} (7-\lambda)\cdot x & -3\cdot y &=0\newline 10\cdot x & +(-4-\lambda)\cdot y &=0. \end{array} \end{eqnarray*} We first eliminate the variable \(y\) from the second equation. This can be accomplished by multiplying the first equation by \(-\frac{-4-\lambda}{-3}\) and adding to the second. We obtain \[\left( \lambda^2 -3\cdot\lambda 2\right)\cdot\frac{x}{-3}=0.\] The non-trivial solution exists only when \(\lambda^2 -3\cdot\lambda 2=0\) or, when \(\lambda\in\{2,1\}\). These are the eigenvalues of the matrix \(A\).

If \(\lambda= 2\), we obtain the eigenvector \(\left[\begin{array}{c}3\newline5\end{array}\right]\).

For \(\lambda= 1\), the corresponding eigenvector is \(\left[\begin{array}{c}1\newline2\end{array}\right]\).

Hence, \(w_1\) and \(w_3\) are the eigenvectors.

We want to find all vectors \(w=\left[\begin{array}{c} x\newline y\newline z\end{array}\right]\) such that \(Aw=\lambda w\) for some real number \(\lambda\).

The given equation is equivalent to the system: \begin{eqnarray*} \begin{array}{rrrl} (2-\lambda)\cdot x & -3\cdot y & 5\cdot z &=0\newline -1\cdot x & +(10-\lambda)\cdot y & -14\cdot z &=0\newline -2\cdot x& 4\cdot y & +(-6-\lambda)\cdot z &=0. \end{array} \end{eqnarray*} We first eliminate the variable \(z\) from the second and third equation. This gives us the system: \begin{eqnarray*} \begin{array}{rrl} \left(23 -14\cdot\lambda\right)x& + \left(8 -5 \cdot\lambda\right)y&=0\newline \left(2 -4\cdot \lambda-\lambda^2\right)x &+ \left(2 -3\cdot \lambda\right)y&=0. \end{array} \end{eqnarray*} Consider first the case when \(\lambda=\frac{8}{5}\). Then \(23 -14\cdot \lambda \neq 0\) and \(2 -3\cdot \lambda\neq 0\). Hence the only solution is \(x=y=z=0\) which is not a non-trivial eigenvector.

Thus we may assume \(\lambda\neq\frac{8}{5}\). We eliminate the \(y\) variable from the second equation by multiplying the first equation by \(-(2 -3\cdot \lambda)\) and adding it to the equation obtained by multiplying the second by \((8 -5\cdot \lambda)\). We obtain: \[\left(-30 55\lambda -30\lambda^2 5\cdot\lambda^3\right)x=0.\] The non-trivial solutions are when \(\lambda\in\left\{3, 2, 1\right\}\). These are the eigenvalues of the matrix \(A\).

If \(\lambda= 3\), we obtain the eigenvector \(\displaystyle\left[\begin{array}{c}-7\newline19\newline10\end{array}\right]\).

For \(\lambda= 2\), the corresponding eigenvector is \(\displaystyle\left[\begin{array}{c}-2\newline5\newline3\end{array}\right]\).

For \(\lambda=1\), the corresponding eigenvector is \(\displaystyle\left[\begin{array}{c}-1\newline3\newline2\end{array}\right]\). Hence, the required vectors are \(w_2\), \(w_3\), and \(w_5\).

• (S1) The statement is false. The matrix \(\displaystyle \left[\begin{array}{cc} 1&1\\-1&1\end{array}\right]\) has no eigenvectors since its characteristic polynomial is \((1-\lambda)^2+1\), which cannot be factored on \(\mathbb R\).

• (S2) The statement is true. Assume that \(u\) is an eigenvector of \(B\) and let \(\lambda\) be the corresponding eigenvalue. Then \(B(Au)=A(Bu)=A(\lambda u)=\lambda A(u)\). Hence \(A(u)\) is an eigenvector of \(B\) corresponding to the eigenvalue \(\lambda\).

• (S3) The statement is false. Let \(\displaystyle A=\left[\begin{array}{cc}2&0\\0&1\end{array}\right]\). Then \(\displaystyle u=\left[\begin{array}{c}1\\0\end{array}\right]\) and \(\displaystyle v=\left[\begin{array}{c}0\\1\end{array}\right]\) are eigenvectors of \(A\). However, \(\displaystyle u-v=\left[\begin{array}{c}1\\-1\end{array}\right]\) is not an eigenvector of \(A\) since \begin{eqnarray*}A(u-v)=\left[\begin{array}{c}2\\-1\end{array}\right].\end{eqnarray*}

Thus, the statement \(S2\) is true, while \(S1\) and \(S3\) are false.