- #1

Dank2

- 213

- 4

Let's say i have the 3x3 matrix

it's eigenvalues are e1 =a, e2 = 0, e3 = 1.

now if a = / = 0, 1

i have 3 distinct eigenvalues and the matrix is surely can be Diagonalizable.

so if i try to solve for the eigenvector for the eigenvalue e1 =a:

after doing elementary operations i get:

now if -a-2b =0, then -b(1-a) =/=0, and vice versa. how do i find the eigenvector for a=/=0,1, if i don't know which one is equal to zero A22 or A23 ?

Code:

```
a 0 0
b 0 0
1 2 1
```

now if a = / = 0, 1

i have 3 distinct eigenvalues and the matrix is surely can be Diagonalizable.

so if i try to solve for the eigenvector for the eigenvalue e1 =a:

Code:

```
0 0 0
b -a 0
1 2 1-a
```

Code:

```
1 2 1-a
0 -a-2b -b(1-a)
0 0 0
```

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