This is given about a matrix A:

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


  • 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?


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


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

Is this significant?

  • 2
    $\begingroup$ $A$ 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 Nov 10 '11 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. $\endgroup$ – JustcallmeDrago Nov 10 '11 at 1:36

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.


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...


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.


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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