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Show that A is diagonalizable? That is, show that A is similar to a diagonal matrix, D, by finding a matrix P such that D= $P^{-1}AP$. Show all your work.

I already found the eigenvalues and eigenvector for matrix A which is

$$ A = \begin{bmatrix} 2 & -4 \\ 3 & 10 \\ \end{bmatrix} $$ The eigenvalues I found were 8 and 4. When I plugged them back into the matrix I found that each eigenvalue has infinitely many eigenvectors. I just picked a value of y to find a values of x to get just one eigenvector.

Now, I'm just wondering how to find out that it is diagonalizable. Is it even possible since I have infinitely many eigenvectors for each eigenvalue?

Ayoshna
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4 Answers4

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Once you have an eigenvector $v$ with respect to the eigenvalue $\lambda$, all non-zero multiples of $v$ will also be eigenvectors with respect to the eigenvalue $\lambda$; so you are observing a natural phenomenon.

You should know that eigenvectors relative to distinct eigenvalues are linearly independent. Hence if you take an eigenvector for $4$ and one for $8$, they will form a basis, with respect to which the matrix will be (obviously!) diagonal.

Andreas Caranti
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Hint:

Just pick one eigenvector for each eigen value. What is the matrix in a basis of eigenvectors?

Bernard
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an eigenvector corresponding to the eigenvalue $4$ is $\pmatrix{2\\-1}$ and the one corresponding to the eigenvalue $8$ is $\pmatrix{-2\\3}.$ so that $$A\pmatrix{2&-3\\-1&3}=\pmatrix{2&-3\\-1&3}\pmatrix{4&0\\0&8} \to \\ A = \frac19 \pmatrix{2&-3\\-1&3}\pmatrix{4&0\\0&8}\pmatrix{3&3\\1&2} \\ A = U\pmatrix{4&0\\0&8}U^{-1}, U = \frac13\pmatrix{2&-3\\-1&3}.$$

abel
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Similar matrices have the same eigenvalues. (See Similar matrices have the same eigenvalues with the same geometric multiplicity).

So, if your matrix $A$ is similar to a diagonal matrix $D$, $D$ has the same eigenvalues 4 and 8. Which means that

$$D= \begin{bmatrix} 4 & 0\\ 0 & 8 \end{bmatrix} $$

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Avi
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  • I solved it and found D but on the diagonal I got 17 and 13. Does it have to be 4, and 8? – Ayoshna May 17 '15 at 18:30
  • Well, as i said similar matrices must have the same eigenvalues, and because the eigenvalues of a diagonal matrix are exactly its elements on the main diagonal, yes it has to 4 and 8. – Avi May 17 '15 at 19:33