Question

A is a rank 1 , 3×3 matrix the null space is 2 Dimensional ,So A has eigenvalue 0 repeated twice.

A= ⎜4 2 4 ⎟
⎟ 2 1 2 ⎟
⎟ 4 2 4 ⎟
How can you say by looking at this matrix , that it has repeated eigenvalue 0 ??

Answer 1

It's been a while since I've taken linear algebra, so I'm afraid I can't gather much from the given information. However, you can show that 0 is a repeated eigenvalue by computation:
\[\begin{align*}\begin{vmatrix}4-\lambda&2&4\\2&1-\lambda&2\\4&2&4-\lambda\end{vmatrix}&=0\\\\\\
(4-\lambda)\begin{vmatrix}1-\lambda&2\\2&4-\lambda\end{vmatrix}-2\begin{vmatrix}2&4\\2&4-\lambda\end{vmatrix}+4\begin{vmatrix}2&1-\lambda\\4&2\end{vmatrix}&=0\\\\
(4-\lambda)\bigg((1-\lambda)(4-\lambda)-4\bigg)-2\bigg(2(4-\lambda)-8\bigg)+4\bigg(4-4(1-\lambda)\bigg)&=0\\\\
-\lambda^2(\lambda-9)&=0
\end{align*}\]

Answer 2

Yes , but Prof. Strang just looked at the matrix and said it would have 2 zeros since it has a 2 dimensional null space , I am wondering someone would give a proof for that .

Answer 3

the first column equals the 3rd column (so they are dependent)
the 2nd column is ½ the first column, so they are dependent.

there is only 1 independent column
and 2 dependent columns (hence two 0 eigenvalues)

Answer 4

@phi, that's using the rank-nullity theorem right?

Answer 5

yes, it is rank 1. so with 3 dimensions, that leaves 2 dimensions for the null space

Answer 6

rank 1. so with 3 dimensions, that leaves 2 dimensions for the null space .
, okay ,but how does it tell it has 0,0 eigenvalue .

please give me link or something so that i can see

Answer 7

the eigenvectors are found by finding the null space of the matrix
\[ A - \lambda I\]
when λ=0, that is equivalent to finding the null space of A
we know we will find two vectors (null space is 2 dimensions)
λ=0 for either of these eigenvectors.
i.e. we have two eigenvalues = 0

Answer 8

thank you phi , I am thinking about your reply friend

Answer 9

Since we know that each lambda contributes atleast one independent eigenvector ,
Now ,phi what if a matrix A have more than n eigenvector ??
how to prove it is not possible ??