# Showing tA matrix to be nonsingular

gordonjh

## Homework Statement

Show that if A ε Mnxn is nonsingular and t ≠ 0, then tA is nonsingular and
(tA)-1 = (1/t)A-1.

## The Attempt at a Solution

I need to show an intense proof of this statement. Although I can grasp the concept in my head, I am unsure as to the mathematical reasons and theorems that prove this true.

## Answers and Replies

Mentor
For starters, what's the determinant of A? Of tA?

gordonjh
Although I understand the relevance of determinants to this particular problem, we have not yet reached that section in the book and therefore are not allowed to use determinants as any part of the proof.

Mentor
OK, since A is nonsingular, it has an inverse, right?

What do you get if you multiply tA and (1/t)A-1?

gordonjh
tA(1/t)A-1=In.

So by taking (tA)-1 = (1/t)A-1 and multiplying both sides by (tA):

(tA)-1(tA) = (tA)(1/t)A-1
(tA)-1(tA) = In

which proves that tA is nonsingular. Additionally, this proves the equality, correct? This is all I need to prove these two statements?

Staff Emeritus
Gold Member
tA(1/t)A-1=In.

So by taking (tA)-1 = (1/t)A-1 and multiplying both sides by (tA):

(tA)-1(tA) = (tA)(1/t)A-1
(tA)-1(tA) = In

which proves that tA is nonsingular. Additionally, this proves the equality, correct? This is all I need to prove these two statements?
You either have it exactly right, or you are confused some. Can you give some more detail on the logic of your proof?

gordonjh
No I am not confused; I suppose I was just expecting more complexity in the proof.

So as I understand it, using the given equality and multiplying both sides by (tA), which would be the inverse of (tA)-1, I receive:

(tA)-1(tA)=(tA)(1/t)A-1
(tA)-1(tA)=(A)(A-1) since t does not equal 0 so t cancels and
(tA)-1(tA)=In since A is invertible and thus A x A-1=In

Therefore, since a matrix, in this case (tA), is nonsingular if the product with its inverse =In, this proves tA to be nonsingular. While proving tA nonsingular, the equation was also proven correct.

Mentor
No I am not confused; I suppose I was just expecting more complexity in the proof.

So as I understand it, using the given equality and multiplying both sides by (tA), which would be the inverse of (tA)-1, I receive:
The equation is not given: it's one of the things you have to establish (the other being that tA is invertible).
(tA)-1(tA)=(tA)(1/t)A-1
(tA)-1(tA)=(A)(A-1) since t does not equal 0 so t cancels and
(tA)-1(tA)=In since A is invertible and thus A x A-1=In
Therefore, since a matrix, in this case (tA), is nonsingular if the product with its inverse =In, this proves tA to be nonsingular. While proving tA nonsingular, the equation was also proven correct.

gordonjh
Okay now I have to admit I am confused. So if I start with (tA)(1/t)(A-1), solving this gives me the In. But from here, since as you say I'm not given the equation I'm trying to prove, I don't know what I can do with this information.

Mentor
If you establish that A has an inverse, and then show that (tA)(1/t)(A-1) = I, then what does that say about (1/t)(A-1)?

gordonjh
(1/t)(A-1) is the inverse of (tA)?

Mentor
Yes. In other words, (tA)-1 = (1/t)A-1.

Notice that we didn't use this equation; we established it.

gordonjh
Ah I see now. So not only does this establish the invertibility of tA, but it also then can be used to prove the equation. Thanks so much Mark44!