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My professor gives me the theorem on non- diagonalisable matrices:

Let a matrix the $A \in M_{n\times n}(\mathbb{R}).$

$A$ has $k$ independent eigen vectors $\Leftrightarrow$ A is similar to $$ \begin{pmatrix} \Lambda & B \\ 0 & C \end{pmatrix} $$ where $$\Lambda= diag(\lambda_1,\lambda_2.......,\lambda_k)= \left[ {\begin{array}{cccc} \lambda_{1} & 0 & \cdots & 0\\ 0 & \lambda_{2} & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda_{k}\\ \end{array} } \right], $$ and $A$ is diagonalisable $\Leftrightarrow$ $k =n$

I have an couple of questions

(1) would you give me an example, by which above all conditions of the theorem are satisfied?

(2) Also I want to visualize the matrix $$ \begin{pmatrix} \Lambda & B \\ 0 & C \end{pmatrix}, $$ when we put the values of $\Lambda$?

(3) why and when $A$ is diagonalisable $\Leftrightarrow$ $k =n$, give me an example.

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  • $\begingroup$ $\Lambda = B=C =1$ is the simplest possible matrix like this. $\endgroup$
    – Paul
    Apr 22 at 14:28
  • $\begingroup$ @Paul if I put $\Lambda$ equal to $2 \times 2$ matrix, then how it looks? $\endgroup$
    – David
    Apr 22 at 14:35

2 Answers 2

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  1. Consider the matrix $$ \begin{pmatrix}1 &0 & 0 &0 \\ 0 & 1 & 0 & 0\\ 0 & 0& 2& 1 \\ 0 & 0 & 0& 2\end{pmatrix}. $$

This matrix has three eigenvectors $e_1$, $e_2$, $e_3$ corresponding to eigenvalues $1$, $1$, and $2$, respectively. In this case, $k = 3 < 4 = n$.

  1. Not sure what you mean by this.

  2. If $k=n$, then $A$ is similar to an $n\times n$ diagonal matrix $\Lambda$ so $A$ is diagonal. As an example, take the previous matrix with the $(3,4)$ entry changed from $1$ to $0$.

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  • $\begingroup$ In point2 I mean if I put diagonal $2 \times 2$ matrix $$ \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} $$ in $$ \begin{pmatrix} \Lambda & B \\ 0 & C \end{pmatrix} $$, then how it looks? I want to see matrix after plugging. $\endgroup$
    – David
    Apr 22 at 15:48
  • $\begingroup$ $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}$? $\endgroup$
    – whpowell96
    Apr 22 at 15:52
  • $\begingroup$ $\Lambda$=$$ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$ ? B=C=1 for your above matrix? $\endgroup$
    – David
    Apr 22 at 15:56
  • $\begingroup$ $B = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ $\endgroup$
    – whpowell96
    Apr 22 at 15:57
  • $\begingroup$ Now I understand, but in point 1 , why your matrix has 3 eigen values and vectors, it should be 4 because it is 4×4 matrix? $\endgroup$
    – David
    Apr 22 at 16:00
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To answer question 2: take $$ \Lambda = \pmatrix{\lambda_1&0\\0&\lambda_2}, \quad C = \pmatrix{1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9}, \quad B = \pmatrix{1 & 0 & -1\\ -2 & -3 & -4} $$ Then the block matrix described is the $5 \times 5$ matrix given by $$ \pmatrix{\Lambda & B\\0&C} = \left( \begin{array}{cc|ccc} \lambda_1 & 0 & 1 & 0 & -1\\ 0 & \lambda_2 & -2 & -3 & -4\\ \hline 0&0&1&2&3\\ 0&0&4&5&6\\ 0&0&7&8&9 \end{array} \right). $$ The lines don't change anything about the properties of the matrix; I've just placed them there to make it clearer where $\Lambda, B, C$ end up within the partitioned matrix.

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