# Two related questions about matrix algebra

1. Feb 17, 2013

### kwal0203

1. The problem statement, all variables and given/known data

a.) If A is an 'n x n' matrix and X is an 'n x 1' nonzero column matrix with

AX = 0

show, by assuming the contrary, that det(A) = 0

b.) Using the answer in 'a' show that the scalar equation which gives the values of λ that satisfy the matrix equation AX = λIX is:

det(A - λI) = 0

2. The attempt at a solution

a.) If det(A) ≠ 0 then A^-1 exists.

X = A^-1 x (AX) = A^-1 x (0) = 0

This is a contradiction because x nonzero so det(A) = 0.... this bit I understand however the next part

b.) AX = λIX -> X(A - λI) = 0

X ≠ 0 so for the equation to be true A - λI = 0

I'm not sure how to apply the first result to the second question? It's from Bimore & Davies Calculus: Concepts and Methods book.

Any help would be appreciated!

2. Feb 17, 2013

### SqueeSpleen

X ≠ 0, does not implies that A - λI = 0
For example
$\begin{pmatrix} 2 &0 \\ 0 &1 \end{pmatrix}$
It's clear that it's eigenvalues are 1 and 2, but...
A-2I=
$\begin{pmatrix} 0 &0 \\ 0 &-1 \end{pmatrix}$
And if we take X=
\begin{pmatrix}
1\\
0
\end{pmatrix}
Then A.X=0
I think, your mistake it's that that it's some X (more precisely, the eigenvectors that the matrix has), but not to all X, otherwhise the matrix has to be λI and thus, A-λI=0 is true, but it isn't necessary.

I think you're pretty close to the answer, you have to think what happens when there's some X such as:
AX = λIX
Then, as you wrote:
(A-λI)X=0

Last edited: Feb 17, 2013
3. Feb 17, 2013

### kwal0203

1.) AX = 0 .... Det(A)≠ 0 -> A^(-1) exists Edit: I had X not equal to 0

X = A^(-1) x AX = A^(-1) x 0 = 0 so det(A) = 0

2.) AX = λIX -> X(A - λI) = 0 .... Let B = A - λI .... If det(B) not equal to 0 then:

X = B^(-1) x BX = B^(-1) x 0 = 0

so det(B) = 0 -> det(A - λI) = 0????

Last edited: Feb 17, 2013
4. Feb 17, 2013

### SqueeSpleen

I recommend you to use dots instead x to the matrix product.
And, while I understand what you tried to say here:
AX = λIX -> X(A - λI)
X is a column vector, so you can only use the product of the square matrix with the X at the right (You can lose points for doing this).

Last edited: Feb 17, 2013
5. Feb 17, 2013

### kwal0203

Thanks for the input! Not sure I completely follow what you say here.

Is this incorrect?

AX = λIX -> AX - λIX = 0 -> X(A - λI) = 0?

Edit: oh great thanks!

6. Feb 17, 2013

### SqueeSpleen

The proof of b) is right, but you don't need to repeat the proof, you can just say that:
(A - λI)X=0 them by a), det(A - λI)=0

AX = λIX -> AX - λIX = 0 -> X(A - λI) = 0
Again:
AX = λIX -> AX - λIX = 0 -> (A - λI)X = 0
:P
And yes, it's correct because the matrix product to right (to left) is distributive with the matrix addition.

Last edited: Feb 17, 2013
7. Feb 17, 2013

### voko

No, this is not correct. AX = λIX -> (A - λI)X = 0. Which is not the same as X(A - λI) = 0.

Now, let (A - λI) = A', so we have A'X = 0. Can you apply part (a) to this?

8. Feb 17, 2013

### kwal0203

Oh right you guys are saying:

X(A-λI) ≠ (A-λI)X

Yeah I just have a habit to put the X in front of brackets all the time. You think I would lose marks for that?

I guess technically I should.

9. Feb 17, 2013

### voko

It is not just that X(A-λI) ≠ (A-λI)X, it is much worse. When X is a column, and A is a matrix, XA simply does not make sense, it is not defined. It is a pretty severe mistake to make, but it is also very easily avoidable: do not change the order of entities in a matrix product, matrix multiplication is not commutative, except in very special cases.