# Theorem on non-diagonalisable matrix [closed]

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.

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

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 $$0$$.

• 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. Apr 22 at 15:48
• $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}$? Apr 22 at 15:52
• $\Lambda$=$$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ ? B=C=1 for your above matrix? Apr 22 at 15:56
• $B = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ Apr 22 at 15:57
• 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? Apr 22 at 16:00

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.