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This is given about a matrix A:

$A \begin{bmatrix} 1\\2\\4 \end{bmatrix} =9 \begin{bmatrix} 1\\1\\1 \end{bmatrix} , A \begin{bmatrix} 1\\−3\\9 \end{bmatrix} =2 \begin{bmatrix} 1\\−3\\9 \end{bmatrix} , A \begin{bmatrix} 1\\1\\1 \end{bmatrix} =3 \begin{bmatrix} 1\\1\\1 \end{bmatrix}$

Find:

  • Eigenvalues & Eigenvectors of A.
  • Is A invertible?
  • Is A diagonalizable?

I'm pretty sure that only $2$ & $3$ are eigenvalues and only $\begin{bmatrix} 1\\−3\\9 \end{bmatrix}$ & $\begin{bmatrix} 1\\1\\1 \end{bmatrix}$ are eigenvectors (at least from the information given). Since the 2nd and 3rd identities given are in the form $Ax = {\lambda}x$.

A will be diagonalizable if and only if the eigenvectors are linearly independant, which they are, so Yes, A is diagonalizable

I have no idea how to find if the matrix is invertible.

Is my thinking correct? And how do I know if A is invertible?


Edit:

As I look at this further I see: $3A\begin{bmatrix} 1\\1\\1 \end{bmatrix} = 3*3\begin{bmatrix} 1\\1\\1 \end{bmatrix} = 9\begin{bmatrix} 1\\1\\1 \end{bmatrix} =A\begin{bmatrix} 1\\2\\4 \end{bmatrix}$

so

$A\begin{bmatrix} 3\\3\\3 \end{bmatrix} = A\begin{bmatrix} 1\\2\\4 \end{bmatrix}$

Is this significant?

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    $\begingroup$ $A$ will be diagonalizable if you can find $n$ linearly independent eigenvectors, where $A$ is $n \times n$. Only two independent vectors are not enough for this example. Indeed, if there are only too independent eigenvectors, what does this imply about diagonalizability? $\endgroup$
    – Shaun Ault
    Commented Nov 10, 2011 at 1:17
  • $\begingroup$ @ShaunAult: Thanks. I added some stuff to the end that I found after looking at the problem more, but I don't know if it will help me or not. $\endgroup$ Commented Nov 10, 2011 at 1:36

4 Answers 4

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As you found above, $$ A \left[\begin{array}{c} 3 \\ 3 \\ 3 \end{array}\right] = A \left[\begin{array}{c} 1 \\ 2 \\ 4 \end{array}\right] $$ Subtracting and factoring out $A$, we find $$ A \left[\begin{array}{c} 2 \\ 1 \\ -1 \end{array}\right] = 0$$ But this means there's a nontrivial vector in the nullspace of $A$, implying another eigenvalue...

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You don't need to recreate the matrix to answer the questions (although, yes, you can reconstruct it). With a little more work, as Shaun Alt showed, you see that 0 is an eigenvalue corresponding to $\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}$ (you are given the other two eigenvalues and eigenvectors as you know). So, now you know all 3 eigenvalues and all 3 eigenvectors. Thus, your eigenvalues are 0, 2, 3. They are distinct. Any matrix with distinct eigenvalues is diagonalizable and the diagonal matrices that A is similar to have 0, 2, and 3 (the eigenvalues) on the diagonal. If you have seen the Jordan canonical form, this is easy to see because all blocks can only be 1 by 1 since each eigenvalue is multiplicity 1. And, by the way, we know each is multiplicity 1 because there can only be 3 total eigenvalues, counting multiplicity, for a 3 by 3 matrix. One of the 6 diagonal matrices that $A$ is similar to would be

$$S^{-1} A S = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}, \qquad \text{for some invertible 3 by 3 matrix } S$$

is not invertible because it has a column of 0s. Since A is similar to this, A also can not be invertible.

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The vectors you gave which A acts on are linearly independent, and so they form a basis for three dimensional space. If you know how a matrix acts on a basis, you know how it acts on any vector, (in particular the standard basis) and so you can actually recreate the matrix. Then the remaining questions are straightforward.

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According to the information you have, 0, 2 and 3 are eigenvalues. You have already given the eigenvectors for 2 and 3. As pointed out by Shaun Ault, $\begin{bmatrix}2\\ 1\\ -1\end{bmatrix}$ is an eigenvector for $0$. It is not invertible because 0 is an eigenvalue, i.e. its kernel is not trivial. Eigenvectors corresponding to distinct eigenvalues are linearly independent and there are 3 distinct eigenvalues. So, there are three linearly independent eigenvectors. So, you have a basis consisting of eigenvectors. Hence, A is diagonalisable.

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