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For each of the following matrices $A$, find an invertible matrix $P$ over $C$ such that $P^{-1}AP$ is upper triangular:

$$A = \begin{bmatrix}4 & 1\\-1 & 2\end{bmatrix} \quad \text{ and } \quad A = \begin{bmatrix}1 & 1 & 1\\1 & 1 & 1\\0 & 0 & 0\end{bmatrix}.$$

For the first matrix, I calculate the eigenvalue to be $3$, with algebraic multiplicity $2$. However, I can only find one eigenvector for it, namely $\begin{bmatrix}1\\-1\end{bmatrix}$.

The second has two eigenvalues, $0$ (multiplicity $2$) and $2$ (multiplicity $1$) But again, for the eigenvalue $0$, I can't find two linearly independent eigenvectors.

So, how do I proceed to find $P$ in these cases?

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    $\begingroup$ relevant: Jordan canonical form, which for the first matrix is $\left(\begin{smallmatrix}3&1\\0&3\end{smallmatrix}\right)$. $\endgroup$ – vadim123 May 12 '13 at 20:37
  • $\begingroup$ @GitGud I'm new to Latex, so don't know how to write it. But I calculated the char poly to ge the eigenvalue. I then did:A-(eigenvalue*I)*v = 0, where v= eigenvector $\endgroup$ – Sapph May 12 '13 at 20:45
  • $\begingroup$ @vadim123 Thanks, I didn't know about the JCF. From what I see, it can't be applied to the second matrix as the (1,3)th entry of A isn't 0. Am I correct? $\endgroup$ – Sapph May 12 '13 at 20:46
  • $\begingroup$ every matrix has a JCF, there will just be multiple blocks. Wolframalpha can calculate this for you like this. $\endgroup$ – vadim123 May 12 '13 at 20:48
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For the first matrix, your work is all correct.

We need to find a generalized eigenvector.

One approach to this (did you learn why in class), is to setup:

$$[A - \lambda I]v_2 = v_1$$

I am going to write the eigenvector with the signs swapped for the first one.

We have $v_1 = (-1, 1)$, so we would get:

$$[A -\lambda I]v_2 = v_1 \rightarrow \begin{bmatrix}1 & 1\\-1 & -1\end{bmatrix}v_2 = \begin{bmatrix}-1\\1\end{bmatrix}$$

Solving this (use RREF), we get a second linearly independent eigenvector of:

$$v_2 = (-1, 0)$$

The matrix $P$ is formed as linear combination of the eigenvectors as $[v_1 | v_2]$, so we get:

$$P = [v_1 | v_2] = \begin{bmatrix}-1 & -1\\1 & 0\end{bmatrix}$$

This will lead to the Jordan Normal Form for the upper triangular matrix as:

$$J = P^{-1}AP = \begin{bmatrix}3& 1\\0 & 3\end{bmatrix}$$

Also note, there are other methods available and sometimes required over the method outlined above.

Can you use the method above for your second problem and see if it bears fruit?

For example, we find two eigenvalues of $\lambda_1 = 2 and \lambda_{2,3} = 0$. For the $\lambda_{2,3} eigenvalue, we will have a row-reduced-echelon-form of:

$$a+b+c = 0$$

We need two linearly independent choices and have a lot of free variables to choose from. For example, we can choose:

$$v_2 = (-1, 0, 1), v_3 = (-1, 1, 0)$$

For the $\lambda_1=2$ eigenvalue, we form:

$$[A-\lambda_1 I]v1 = 0$$.

After RREF that matrix, we arrive at: $v_1 = (1,1,0)$

Recall, you have to do this process over for EACH eigenvalue!

So, to summarize, we should arrive at (no generalized eigenvectors needed because of geometric and algebraic multiplicity):

$$\lambda_1 = 2, v_1 = (1, 1, 0)$$

$$\lambda_2 = 0, v_2 = (-1, 0, 1)$$

$$\lambda_3 = 0, v_3 = (-1, 1, 0)$$

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  • $\begingroup$ $$\begin{bmatrix}1 & 1 & 1\\1 & 1 & 1\\0 & 0 & 0\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} $$ gives us eigenvector $$\begin{bmatrix}1\\1\\1\end{bmatrix}$$ $\endgroup$ – Sapph May 12 '13 at 21:06
  • $\begingroup$ And so I solve $$\begin{bmatrix}1 & 1 & 1\\1 & 1 & 1\\0 & 0 & 0\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}1\\1\\1\end{bmatrix} $$ Which gives a = 1-b-c , thus we get eigenvalues: $$\begin{bmatrix}1\\0\\0\end{bmatrix} , \begin{bmatrix}-1\\1\\0\end{bmatrix} , \begin{bmatrix}-1\\0\\1\end{bmatrix}$$ Is that correct? $\endgroup$ – Sapph May 12 '13 at 21:07
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    $\begingroup$ Did you mean eigenvectors? $\endgroup$ – Amzoti May 12 '13 at 21:08
  • $\begingroup$ Sorry, This is my attempt at the 3x3! $\endgroup$ – Sapph May 12 '13 at 21:08
  • $\begingroup$ See my update for the eigenvalues, eigenvectors. They need not be unique. $\endgroup$ – Amzoti May 12 '13 at 21:10
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Here is an alternative approach. Note that you are not required to find the Jordan form of $A$. So, when $A$ is $n\times n$, you only need to find one eigenvector of length $n$.

More specifically, suppose $(\lambda_1,v_1)$ is an eigenpair for $A$. Extend $v_1$ to a basis $\{v_1,\ldots,v_n\}$ of $\mathbb{C}^n$ and put the basis vectors together to form an invertible matrix $P_n$. Then $$ AP_n=(\lambda_1v_1,\,Av_2,\,\ldots,\,Av_n) =P_n(\lambda_1e_1,\,P^{-1}Av_2,\,\ldots,\,P^{-1}Av_n) $$ where $e_1=(1,0,\ldots,0)^T$. In other words, $$ P_n^{-1}AP_n=\pmatrix{ \lambda_1&\ast&\ast&\cdots&\ast\\ 0&\ast&\ast&\cdots&\ast\\ 0&\ast&\ast&\cdots&\ast\\ \vdots&\vdots&\vdots&&\vdots\\ 0&\ast&\ast&\cdots&\ast\\ } =\pmatrix{\lambda_1&u^T\\ 0&B} $$ for some $(n-1)$-vector $u$ and some matrix $B$ of order $n-1$. So, if $(\lambda_2,w)$ is an eigenpair of $B$ and $P_{n-1}$ is an invertible matrix whose first column is $w$, then $$ \pmatrix{1\\ &P_{n-1}^{-1}}(P_n^{-1}AP_n)\pmatrix{1\\ &P_{n-1}} =\pmatrix{ \lambda_1&\ast&\ast&\cdots&\ast\\ 0&\lambda_2&\ast&\cdots&\ast\\ 0&0&\ast&\cdots&\ast\\ \vdots&\vdots&\vdots&&\vdots\\ 0&0&\ast&\cdots&\ast\\ }. $$ Proceed recursively, you get an upper triangular matrix.

For your two $A$s, it happens (for different reasons) that you only need one shot each to get a triangular matrix. Take your second $A$ (which is $3\times3$) as an example. $(1,1,0)^T$ is an eigenvector of $A$ corresponding to the eigenvalue $2$. So, if we put $$ P_3=\pmatrix{1&\ast&\ast\\ 1&\ast&\ast\\ 0&\ast&\ast} $$ where the second and third columns are arbitrary (as long as $P_3$ is invertible), we have $$ P_3^{-1}AP_3 = \pmatrix{2&\ast&\ast\\ 0&0&0\\ 0&0&0}. $$ (Exercise: For each of the two $A$s in question, why only one shot is enough?)

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