# Is the matrix diagonalizable for all values of t?

For tâˆˆR, let $A_t = \left( \begin{array}{ccc} t & 1 & 1 \\ 1 & t & 1 \\ 1 & 1 & t \end{array} \right)$. Find the Eigenvalues and Eigenvectors. Is $A_t$ diagonalizable for all t?

So I am thinking of approaching it like this. First, I find the characteristic polynomial of $A_t$ and use that to find the eigenvalues and their corresponding eigenvectors. Then I check if those eigenvectors are linearly independent for all t. If they are linearly independent, then $A_t$ is diagonalizable for all t, and if they’re not, $A_t$ is not diagonalizable for all t.

Should I pursue this problem like this? Or is there a more efficient method? Thanks in advance!

#### Solutions Collecting From Web of "Is the matrix diagonalizable for all values of t?"

Hints.

1. Explain why $t-1$ is an eigenvalue. What can you say about its multiplicity?
2. The sum of all three eigenvalues is equal to the trace of the matrix.

Then find eigenvectors and proceed as you have proposed.

Suggestion: never calculate a characteristic polynomial unless you really really REALLY have to ğŸ™‚

Your approach should work, but a more efficient method would be to note that $A_t = A_0 + t I_3$. Adding a multiple of the identity to a matrix doesn’t change its eigenvectors, so you can just find the eigenvectors of $A_0$, and these will also be eigenvectors of $A_t$. You can then apply $A_t$ to them to determine the eigenvalues.

Whether a matrix is diagonalisable or not does not change when a multiple of the identity is added (think of it after a possible change of coordinates making it diagonal: since the identity matrix unchanged by change of coordinates, you are just adding a multiple of the identity to the diagonal form, which will keep it diagonal).

So your question boils down to whether the matrix with all entries$~1$ (obtained for $t=1$) is diagonalisable or not: I think you can answer that question easily.

Let $v_{1} = \begin{pmatrix} 1 \\1 \\1 \end{pmatrix}$, $v_{2} = \begin{pmatrix} 0 \\1 \\-1 \end{pmatrix}$, $v_{3} = \begin{pmatrix} -1 \\0 \\1 \end{pmatrix}$.

Let $B$ be the $3 \times 3$ matrix whose columns are $v_{1}, v_{2}, v_{3}$, then $\det B = 3 \neq 0$ so that the set $\{v_{1}, v_{2}, v_{3}\}$ is linearly independent, but then it must be a basis of $V = \mathbb{R}^{3}$.

We also know that $A_{t}v_{1} = (2 +t)v_{1}, A_{t}v_{2} = (t-1)v_{2}, A_{t}v_{3} = (t-1)v_{3}$.

Hence, we have found a basis of $V = \mathbb{R}^{3}$ consisting of eigenvectors of $A_{t}$, but this is one way of characterizing diagonalizability of a matrix (a linear operator) and so $A_{t}$ is diagonalizable as it was promised.

Edit: Building vectors in a similar manner works when the dimension is change from 3 to $n$. This is one easy example to remember as a counterexample to a matrix can be diagonalizable even if it has repeated eigenvalues [however when the eigenvalues are distinct we have diagonalizability and if we are to use this test we have to calculate the characteristic polynomial]