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I have stumbled across a very old exam question from my linear algebra course and the solutions are not available. I was wondering if my working/logic is correct and if any improvements can be made.

Let $\ B=\begin{pmatrix} 1 & -1 & 4 \\ 6 & -7 & 2 \\ -3 & 1 & -6 \end{pmatrix}\ $ and $\ \vec{v_1}=\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.

Part (i): Calculate $(B+5I)\vec{v_1}$ and $(B+5I)^2\vec{v_1}$.

Using simply matrix multiplication, I calculated $(B+5I)\vec{v_1}=\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and $(B+5I)^2\vec{v_1}=\vec{0}$.

Part (ii): Find all eigenvalues and eigenvectors of B.

From our calculations in part (i), we can see that $\ \lambda=-5,-5$ are two of the eigenvalues, with corresponding eigenvector $\ \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and generalised eigenvector $\ \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$. From the trace of B, the remaining eigenvalue is $\lambda=-2$ with corresponding eigenvector $\begin{pmatrix} -2 \\ -2 \\ 1 \end{pmatrix}$.

Part (iii): Giving reasons, write down a basis for the generalised eigenspace $GE_{-5}$ of B.

I am unsure of this part of the question. If I had to guess, I would say a basis for $GE_{-5}=\left\{\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix},\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\right\}$. Is this correct?

Part (iv): Solve the initial value problem $\vec{y'}=B\vec{y}$ where $\ B=\begin{pmatrix} 1 & -1 & 4 \\ 6 & -7 & 2 \\ -3 & 1 & -6 \end{pmatrix}\ $ and $\vec{y}(0)=\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.

Well, the solution to this IVP will take the form $\vec{y}=e^{tB}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$. I noticed that $\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ was a generalised eigenvector and hence we can write $\vec{y}=e^{-5t}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}+e^{-5t}\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}t$. Thus $\vec{y}=e^{-5t}\begin{pmatrix} 1+t \\ 1+2t \\ -1-t \end{pmatrix}$

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  • $\begingroup$ is it possible to provide a solution? I don't see where I am wrong. $\endgroup$
    – user557493
    Jun 3 '18 at 6:53
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    $\begingroup$ @Moo The initial value forces the remaining eigenfunction to go to $0$. $\endgroup$
    – Dylan
    Jun 4 '18 at 4:12
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You can rewrite the matrix in its Jordan normal form $$ \textbf{B} = \textbf{V} \textbf{J} \textbf{V}^{-1} $$

where

$$ \textbf{J} = \begin{bmatrix} -5 & 1 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -2 \end{bmatrix}, \quad \textbf{V} = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -2 \\ -1 & -1 & 1 \end{bmatrix} $$

Applying the matrix exponential gives

$$\vec{y} = e^{t\textbf{B}}\ \vec{y}_0 = \textbf{V} \begin{bmatrix} e^{-5t} & te^{-5t} & 0 \\ 0 & e^{-5t} & 0 \\ 0 & 0 & e^{-2t} \end{bmatrix} \textbf{V}^{-1}\vec{y}_0 $$

Since $\vec{y}_0$ happens to be the second column of $\textbf{V}$, it follows that

$$ \textbf{V} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \vec{y}_0 \implies \textbf{V}^{-1}\vec{y}_0 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $$

and so

$$ \vec{y} = \textbf{V} \begin{bmatrix} te^{-5t} \\ e^{-5t} \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}te^{-5t} + \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}e^{-5t} $$

Your answer is correct.

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  • $\begingroup$ Thank you a lot!!! $\endgroup$
    – user557493
    Jun 4 '18 at 4:32

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