[Math] Show square matrix, then matrix is invertible

linear algebramatrices

Question:
Show that if a square matrix $A$ satisfies the equation $A^2 + 2A + I = 0$, then $A$ must be invertible.

My work: Based on the section I read, I will treat I to be an identity matrix, which is a $1 \times 1$ matrix with a $1$ or as an square matrix with main diagonal is all ones and the rest is zero. I will also treat the $O$ as a zero matrix, which is a matrix with all zeros.

So the question wants me to show that the square matrix $A$ will make the following equation true. Okay so I pick $A$ to be $[-1]$, a $1 \times 1$ matrix with a $-1$ inside. This was out of pure luck.

This makes $A^2 = [1]$. This makes $2A = [-2]$. The identity matrix is $[1]$.

$1 + -2 + 1 = 0$. I satisfied the equation with my choice of $A$ which makes my choice of the matrix $A$ an invertible matrix.

I know matrix $A *$ the inverse of $A$ is the identity matrix.

$[-1] * inverse = [1]$. So the inverse has to be $[-1]$.

So the inverse of $A$ is $A$.

It looks right mathematically speaking.

Anyone can tell me how they would pick the square matrix A because I pick my matrix out of pure luck?

Best Answer

The inverse of $A$ is almost certainly not $A$. You have $A^2+2A=-I$, or $A(A+2I)=-I$. Multiplying by $-1$ you get $$A(-A-2I)=I$$

Hence the inverse of $A$ is $-A-2I$.

Related Solutions

[Math] Proving that a square matrix whose kernel is $\{0\}$ is invertible

I would do it using the rank-nullity theorem and properties of linear maps. If $Ker(A) = {0}$ Then the linear map which the matrix represents is injective and has nullity zero. Hence the rank of the matrix is $n$ (if the matrix is $n\times n$,) which means that it is a surjective map. Since the map is linear and a bijection, it is an isomorphism which means it has an inverse. Hence the matrix must be invertible.

(Of course, to say this rigourously you'd have to explicitly state where the map is from and to and all the rest of it, but if you know a little bit about it it's not hard to do.)

[Math] Show that if $A$ is invertible and $AB = AC$, then $B = C$.

Generally speaking, you can't prove anything by example. You can prove all integers are even if you are allowed to "prove by example". I pick 2, 98, and 600! See? Everything I tried, worked! But that doesn't tell you anything about "all integers". So, it's not enough to show that it works in one or two or a hundred cases.

[On the other hand, working out a few examples can be extremely beneficial in an effort to figure out why some idea X has to be true in all cases, and can lead you to a proof.]

In this case, however, you're looking at an example of why they say that "math is 40% writing 1 in a creative way, and 40% writing zero in a creative way".

The trick with this one is what you might call a "hidden given". They gave us that A is invertible, and they gave us that AB = AC. But what is the hidden given? The hidden given is that B is equal to itself. Very tricky, and not immediately obvious how it helps us.

But this is what we're going to do with it: we know that B = B. We're going to start with that equality, and play around mathematically until it looks like what we want to prove (namely, that B = C).

So, here we go:

B = B

How can we get an A into that equation? There's a reason for us to want to do that. AB is equal to AC. If we can put an A in that equation somehow, we might then be able to get a C in there, and then maybe we can get it to tell us that B = C.

Well, here's where we use the rule that much of math is "creatively writing one". One times B is B--in matrix-ese, that is "IB=B". Your brain has to make the leap that--since we can also write I as A^-1A--this is our ticket to getting the A in there where once we only had the boring fact that B=B.

To be formal, we "know that IB = B, by definition of the identity". And when they say that A is invertible, in formal terms that means that we know that "there exists an A^-1 such that A^-1A = I and AA^-1 = I".

So, in this little equation we get from applying the definition of the identity to "B = B":

IB = B

, we're going to creatively write I as A^-1A.

(A^-1A)B = B

Note that all we've done here is to use the definition of the identity and the definition of an inverse--because of what the identity means, we can write IB=B. And because of what invertible means, we can write A^-1A = I.

Now we're going to use associativity--the fact that matrix multiplication can be grouped however we want. The fact that for any X, Y, and Z, (XY)Z is the same as X(YZ). So, going back to where we were

(A^-1A)B = B

we can regroup that stuff on the left like this:

A^-1(AB) = B

Why would we do that? Because it gives us an (AB) to work with. And they told us something about AB! What did they tell us? That AB=AC. Well, if (AB) is equal to (AC), when we can put that AC right in there for (AB).

A^-1(AC) = B

So, to recap where we are--they told us AB=AC. And they told us that A is invertible. And we took those facts, and used them together with a bunch of properties and definitions: (1) what invertible means (2) definition of the identity (3) matrix multiplication is associative (we an even add (0) that every matrix is equal to itself--"equality is reflexive", is how you say that formally).

Well, looking at A^-1(AC) = B, it's just screaming at us to apply associativity again. Let's group up that A^-1 with its old buddy A again, shall we?

(A^-1A)C = B

Now, then, the definition of an inverse tells us that A^-1A is I. Let's apply that:

IC = B

Now let's apply the definition of identity. That "IC" we've got is just equal to C, so:

C = B

It's beneficial to try to get an idea of where you needed the givens. If A wasn't invertible, what would go wrong?

Well, suppose A was the zero matrix (which is not invertible). If A is the zero matrix, then knowing that AB = AC doesn't necessarily tell you anything about B and C--you could literally put any B and C in there, and the equality would still hold. So you need the fact that A is invertible if you want to go from AB = AC to B = C. Your high school algebra brain just screams at you "Cancel the A's! CANCEL THE A's NOW!". But it was never ok to divide both sides by anything unless you were sure that the "anything" wasn't zero. And in matrix-land, things are a little more complex, but there's a similar rule. If it's not invertible, you don't know if you can "cancel" it or not. And that's really what this exercise is about--it's about learning when you can go from AB = AC to B = C. And the answer to that is, "when A is invertible".

Best Answer

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[Math] Proving that a square matrix whose kernel is $\{0\}$ is invertible

[Math] Show that if $A$ is invertible and $AB = AC$, then $B = C$.

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