Question
Let $A$ be an $n imes n$ (square) matrix. Prove the following two statements:(a) If $A$ is invertible and $A B=0$ for some $n imes n$ matrix $B$, then $B=0$.(b) If $A$ is not invertible, then there exists an $n imes n$ matrix $B$ such that $A B=0$ but $B eq 0$
Let $A$ be an $n imes n$ (square) matrix. Prove the following two statements: (a) If $A$ is invertible and $A B=0$ for some $n imes n$ matrix $B$, then $B=0$. (b) If $A$ is not invertible, then there exists an $n imes n$ matrix $B$ such that $A B=0$ but $B eq 0$
Answers
$\begin{array}{l}{\text { If } A \text { is nonsingular, then }\left(A^{-1}\right)^{-1}=A} \\ {\text { (a) Verify this theorem for } 2 \times 2 \text { matrices. }} \\ {\text { (b) Prove the theorem for any } n \times n \text { matrix. }}\end{array}$
This problem. We want to show that the matrix equation a X equals B has a unique solution off inverse of a times be given a few assumptions. We are assuming that a is in vertebral and it's an n by n matrix. And we're told that B is an N by P. Matrix. So the sizes air going to allow us to do the matrix multiplication that we have. So we want to show that this is true. So we're gonna break this into two parts. First, I want to show that this is indeed a solution, and then we will show that it is a unique solution. So let's start with the first part. Is this really a solution? Well, yes, it is. If I start with my equation A X equals B and I'm going to substitute in my my, uh, supposed solution, we're going to show that it really does work. So I'm going to plug that in for X. So I have a times the inverse of a times B equals B. So I've made that substitution matrices have associative property holds for matrices so I can multiply these, um I can I can switch the order in which I multiplying Aiken, instead of having my parentheses around inverse of a and B, I could move the parentheses to a times the inverse of a well, that is the identity matrix and the identity matrix. Times B is just be So this solution works. It gives me ah, tautology at the end. Now, next question. Is it unique? Well, let's let you be a sorry b A solution. Okay, I'm you is a solution. It's any solution. And if you is a solution, that means eight times you will equal be That's the definition of the solution. Well, now, let's take both of these and multiply by the inverse of a in verse of a times A is the identity matrix, which just equals you. So if you is a solution, it has to be the inverse of a times B. So it is a solution, and it is unique. So this is true
So we have be minus C that multiplies the equal to zero. So since the is in vertebral, we can multiply from the right by the inverse of the so right B minus c the and then the inverse. And that's equal to zero times Demers. Now the product the D in verse is just it. Entity metrics. So this becomes B minus c that multiplies the again. Did he is equal to now Any metrics multiplied by 00 So on the right hand side, we just have zero and now must be bribed by the identity matrix does nothing. So on the left hand side we have B minus c and on the right inside we have zero Now we some ah si on both sides. So that's right, B minus c and then we add plus C on the right. Inside we have zero Blasi Well now minus C plus C is just zero. So are we on the left inside? We just have B and on the right inside something zero does nothing. So we're just left with C, which is what we wanted
For this problem. We have three matrices A, B and C. We've been told to things about them. We've been told that these are in vertebral and that they're square matrices, their end by end, which means that we can multiply them together nicely. We don't have to worry about sizes when we go to multiply these matrices and what we want to show is that their product, a B C, is also in vertebral. So what that would mean is, if I have a b c time some D that gives me the inverse or it gives me the identity matrix. And if I switch them around that also gives me, um, the identity matrix. If both of these air true, then I can say that ABC is in vertebral. So let's begin. Let's take the first one A B C times d equals I. Well, let's try to solve this for de what matrix d will this work for? Well, A is in vertebral so I can multiply both sides by the inverse of a Now remember order is very important when we do matrix multiplication on multiplying this on the left hand side because I wanna multiply that inverse of a times a directly So a universe of a times A is just the identity matrix. Uh huh. And in verse of a times, the identity matrix is the inverse of a And because I is the identity matrix, I can remove that from here that those were equivalent statements. Okay, Now let's multiply both sides by the inverse of B well, over here in verse of B times be its identity matrix, which I can leave off, Which leaves me with C and D Hope so see, times D equals the inverse of b times the inverse of a on I'm gonna do this one more time. Multiply both sides by the inverse of C. That leaves me with just the D on the left hand side. Okay, so in order for this to be true, that would have to be the value of B for that first piece to be true. Now what I want to do is if I come back up here to show you the thes two statements, I want to show that the same D that works here also works here. So I'm going to take the second statement solve for D and show that my two DS are equal. Okay, so here's our second statement de times a times B times C equaling. I I'm going to the exact same process. I'm gonna multiply. But this time I'm going to be on the right hand side because I want to be multiplying C times the inverse of see directly. I need to put them next to each other. So it's got gives me d A B. See, Time's is inverse is just the identity matrix. So I can leave that off the identity matrix times the inverse of C is the inverse of C. Okay, Next, I'm gonna multiply both sides by the inverse of B and then I'll do it once more for a okay, And if you notice the same, be the same. Matrix D is created in both cases. So this truly is this. ABC is indeed in vertebral