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Could anyone only help me to find out which one is diagonalizable?

Myshkin
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  • Do you know the criteria for a $n\times n$ matrix to be diagonalizable? – projectilemotion Apr 18 '17 at 07:39
  • yesssssssssssssssssssssssssssss, if its minimal polynomial factors into linear factors – Myshkin Apr 18 '17 at 07:41
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    Please don’t include the bulk of your question as an image. It’s neither searchable nor accessible to people using screen readers. If you take your own time to do this, then more people might be include to take up their own time answering your questions. – amd Apr 18 '17 at 18:20

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Hint: It's not hard to see that \begin{align} v_1= \begin{bmatrix} 1\\ 1 \\ 0\\ 0 \end{bmatrix}, \ \ v_2= \begin{bmatrix} 1\\ -1 \\ 0\\ 0 \end{bmatrix} \end{align} are eigenvectors with corresponding eigenvalues, $3$ and $1$ respectively. Consider $B-3I$ and $B-I$ and see what you get.

Jacky Chong
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  • Perhaps you meant "it's not hard to check" instead of "to see", because I can't see easily at all how to get those eigenvalues without explicitly calculating the characteristic polynomial... – DonAntonio Apr 18 '17 at 08:54
  • @DonAntonio Because of the zero block at the lower left corner of the matrices we could isolate the upper left corner matrix. It's not hard to see $\begin{bmatrix} 2 & 1\ 1 & 2 \end{bmatrix}$ has eigenvectors $[1, 1]^T$ and $[1, -1]^T$, with corresponding eigenvalues $3, 1$. This is how I constructed the eigenvectors for the original matrix. – Jacky Chong Apr 18 '17 at 20:37