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Let $A$ ne an $n\times n $ upper triangular matrix with complex enries.I have to check the validity of the statement

If $A\neq I$ and if $a_{ii}=1$ $\forall i, 1\le i\le n$, then $A$ is not diagonalizable.

I wanted to know whether my proof(given below) is correct or not?Please provide the correct proof... enter image description here

user729424
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Anwar
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1 Answers1

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The claim that there is no pivot volumn is false.

Consider an explicit example: $A = \begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$

then we have $A-I=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$

The second column is a pivot column.

In general, we know that for your setting, we are going to have positive number of pivot columns.

Siong Thye Goh
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  • :What is the Geometric multiplicity then?What is the number of free variables? – Anwar Feb 12 '20 at 17:10
  • :I think its Geometric multiplicity is 2,because it has 2 free variables(there are two columns without pivots $C_1$ and $C_n$)? – Anwar Feb 12 '20 at 18:09
  • For my particular matrix, let's solve for $(A-I)x=0$, the only condition is $x_2=0$, $x_1$ is the free variable here, it can take any value, $x_2$ must take value $0$, it is not a free variable. The leading non-zero element of $A-I$ appears as the second column, it is not a free variable, There is no leading non-zero eleemnt in the first column, it is a free variable. The geometric multiplicity is equal to $1$ in my example. – Siong Thye Goh Feb 13 '20 at 01:26