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I have a matrix and I need to prove that it's diagonalizable for some values of an variable or not diagonalizable at all.

My thoughts are that that the easiest way to do so is by proving that the characteristic polynomial of the matrix is of the form:

$p(\lambda) = (\lambda - \lambda_1)(\lambda - \lambda_2) \cdots (\lambda - \lambda_K)$ where $\lambda_1, \lambda_2, \ldots, \lambda_K$ are the eigenvalues of the matrix.

In order to find the polynomial though I'm trying to solve the equation $\det (A- \lambda Ι_n) = 0$, where $A$ is my matrix.

That equation seems unsolvable on my $4 \times 4$ matrix which is: \begin{pmatrix} 4 + l &11 &4 &3 + l\\ 11 &13 &5 &8\\ 4 &5 &7 &12\\ 3 + l &8 &12 &18 \end{pmatrix}

My questions are these:

Is the format of my characteristic polynomial correct? Are $\lambda_1,\lambda_2, \ldots, \lambda_K$ supposed to be $A$'s eigenvalues? Is there any simplest way to produce the wanted result?

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1 Answer 1

I'm a bit confused here - is this a trick question, or am I just not seeing something obvious?

The matrix that I see is a real symmetric matrix for all $l$. Hence, it is diagonalizable (even by orthogonal matrices) via the spectral theorem. Calculating the characteristic polynomial seems to be overkill...

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Minor nitpick: For all real $l$, but otherwise, I agree with your point. Since the question is not to find all $l \in \mathbb{C}$ for which the matrix is diagonizable, we are indeed done here. –  Roland Feb 25 at 19:03
    
Sure thing with $l\in\mathbb{R}$, that's of course important. –  Martin Feb 25 at 19:06

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