let-a-be-any-invertible-n-times-n-matrix-show-that-for-mathbfu-mathbfv-in-mathbbrn-the-formula-langl

Question

Let $A$ be any invertible $n 	imes n$ matrix. Show that for $\mathbf{u}, \mathbf{v}$ in $\mathbb{R}^{n},$ the formula $\langle\mathbf{u}, \mathbf{v}angle=(A \mathbf{u}) \cdot(A \mathbf{v})=(A \mathbf{u})^{T}(A \mathbf{v})$ defines an inner product on $\mathbb{R}^{n} .$

Anthony Ramos · Accepted Answer

Hello there. Okay, So for this exercise, let&#x27;s consider first we got a is going to be an end times and matrix in vertebral metrics. Okay? In vertical matrix. Also, we got vectors called you on V on our n. Well, there, Andi, let us define on the inner product for this space as follows. So the inner product Yeah, uh, is defined as taking a you the product with a the Okay, but this is just taking a You just pose and the Okay, So this is how is defined in the product. And we need to show that this is a corresponds to again. I mean, on in their problems. Okay, so we need thio to show that satisfy the actions off animal problem. So let&#x27;s start the first that we need that we need todo to show. Okay. Is that you the Is it supposed to be used? It means that the order off the inner product off the vectors doesn&#x27;t affect. Okay, so let&#x27;s do that. Let&#x27;s start. So you the is equals to take into the by definition off the product A You proposed a the beef. Then we can expand this oppression here, so this becomes u transpose a transpose a on the Okay. So Well, we know that this term here is going to be just a number so we can take the transpose of this expression on. We&#x27;re going toe the same value because this here is going to be in a scaler after evaluating after performing it operation between these between the doctors on the matrices. So this is just a number. So if we take here, the transpose of the expression doesn&#x27;t affect their soul. So when we take here the trust pose we obtained retrace both a Just pose a on you. Okay, so this part here, we can associate these thes terms on what we obtain is that this part is equals to a V transpose on this part here is just a you which corresponds to taking the you. So we have shown the first action. Let&#x27;s continue the other property they will need to show is that Yeah, the second one. Okay, that we need to show Is that given u plus v taking the inner product off? This is equivalent to taking thio distribute distribution. Okay. The stoop activity property to this plus the the Okay, so we need to show this right. Let&#x27;s start by taking the left hand side So the left hand side is equals to take in a times you lost the transpose a w Yeah, This is equals to you love v transpose aid to expose a love you And here we can distributed supposed to each component So this is you transpose plus a transpose 88? No. Yeah. Okay, so here we can distribute the operation off A So we obtained u transpose a transpose waas re transposed a transpose times a love you And this is just UT 88 The view plus Bt 80 a love you this part here we can Well, we can rewrite them. Andi, they become a u A fellow of you Waas a Sorry a V here. Just both here. Just both transpose Hey, value on this is just you w class I&#x27;m sorry. Here the w So the second action is hold a swell. Let&#x27;s continue the third one say that if we multiply by any scaler seat, the inner product is equal to take him to see out off the inner product. See you under where c is just like a real member just on the scaler body. Yeah. Okay, so let&#x27;s show this. So let&#x27;s consider this part. He you love you is a time. See you transpose a lot of you is just taking C See you transpose a transposed a love you here the sea We can take it out because it&#x27;s just a killer value So we&#x27;ll see Hugh transposed a transpose a above you We can regret this and get things See that multiplies a you transpose a love you And here is clear that it is equal to see you enough so we can take out any scale. So this is also hold on. The final action that we need to show is that the inner product off a vector with itself is greater or equal to zero on the equality holds if and only if you is equal to zero. Okay, so let&#x27;s start. But so in this case, we&#x27;re going to define a Yeah, you transpose as a one eight to the A n. Okay, so define it in this way. When we compute this you in the product here you think a you transpose a u But this off the evaluating the This is a product we obtain that is equals toe a square plus a two square plus the plus eight and square. So here, clearly each of these components are going to be either zero or greater than zero, because off the square, they&#x27;re all positive numbers. So these person here is going to be greater or equal than zero. So the first part is checked. Now, the second one, when this is equals to zero, so you is equal to zero. This is equivalent to say that a one square plus a two square plus that a N square is equal to zero. Okay, but this is equivalent to say that each of these components the only possibility to this estimation off Bostic numbers to be equals to zero is that all the elements are zero. So that implies that a one a two is equal to the goes to a n on a lot. These terms are zero. Okay, if all these terms are zero, that means just by what we have to find that a U is equal to zero. But we got that a is an in vertebral matrix. So this implies this is equivalent. To say that you ease equals must be equals to zero. So we have shown that the inner product off you with itself is equals to zero if and only if you is equal to zero and that&#x27;s it, that&#x27;s all. So we have shown that the A yeah, that for any Haymon&#x27;s A matrix A. And in vertical metrics, I can the inner product off two vectors you can be on our n define as a u transpose a the peace I&#x27;m in their product.

Pam Russell · Answer

so to be in vertebral, the inverse of a matrix Times itself has to equal the identity matrix. And so we want to prove that the identity matrix minus a is in vertical. So to do that, we&#x27;re gonna take the inverse of this matrix, multiply it by itself, and we&#x27;re hoping that will get the identity matrix. Now we were given that the inverse of this matrix waas the inverse plus a plus matrix, a squared plus matrix a cute we&#x27;re going to multiply that by our matrix. So, using our multiplication, we will take the identity times the identity that&#x27;s just going to give us an identity, the identity times matrix A. But notice that it is theon position of a so we&#x27;ll get negative a the identity times a And then we&#x27;ll have a times the opposite of itself. And that&#x27;s going to give us a square a square times. The identity will be a square, a squared times matrix a but its opposite that will give us a opposite. A cube, a cube times the identity will give us a cube in a cube times the opposite of a will give us A to the fourth Now we can go ahead and reduce some of this and simplify it down, and we get the identity minus a to the fourth. But we were given in the problem statement that aid to the fourth equaled zero. And so if we substitute that in, we get that is equal to the identity. And we have just proven that the identity matrix minus a is in veritable.

Angelo Rendina · Answer

we want to show that the power off the inverse is the inverse off the power. So let&#x27;s proceed by induction on end, so I find is equal to one. Well, that&#x27;s done. There&#x27;s nothing to show because a inverse is equal to a mers. Now let&#x27;s assume that the proposition is true for some value of n Gayatri golden. One we want to show is that a mers to the power and plus one will want to show the teacher is equal to a to the power when pass one in verse. So let&#x27;s start by writing these as well, a members to the power of end times a inverse just once. Now let&#x27;s use the inductive step right chambers and as eight again in verse, that multiplies a inverse. Now these is just equal to eight times a n in verse where we use the property that the product of the inverse ease off two matrices is the embers off the products in the opposite order. All right, And now, inside the brackets, we can write these as a and plus one, everything humors and a profits done

Angelo Rendina · Answer

So we have these linear transformation tea from a vector space v T o r n, and we have, ah interpret well where the finding. I mean a product on V given by the brother of you against V is by definition to you dot tv. We remember that the dot there is in a product that standard in the product inside our end because the U. N TV are factors in our end and so we just need to check that he&#x27;s in a product against your V defined in the vector space. Big B is actually the product, so we need to check the four actions. So what are we, the 1st 1? Which is the symmetry? So we want to check that V against you is by definition tv dot to you and these by the symmetry off the dot product in rn. This is TV at you. Sorry, not TV and all these by definition you against me. So dysfunction on V is symmetric they only to try that it&#x27;s leaner. So x plus y against a vector V We want to check the disease X against being plus wagons being so we use the definition of the inner product. Well, these in a product t x plus y dot d y. And now again by linearity off T and by the charity off the dot product. In our end, this becomes tx dot tv bluff the y dot TV And now again, by definition, did his ex against v blast. Why against me Now for the third property we need to check it is or more genius So CX against V. It is, by definition, tee off, see axe dot tv. But now both t and dot product are linear. Well, that I&#x27;m a genius. So this becomes C t X dot tv and again, these by definition is c times X against me. And finally, we need to check that the dot product, while the inner product is positive definite off. We need to change X against ex. He&#x27;s got their equal to zero so exciting. Sex is by definition, tx dot tx. And now because dot is another product on our end, we know that this is definitely great unequalled in zero. And also this can only be zero if t x zero because the season in the product So did your brother have something against itself can be zero only director itself. Zero. But now remember that tea is want one. So the only victory gets within the vector in V that gets maps to the zero vector. Our end zero is zero factor in we and that means that X must be zero. And therefore we have checked for actions showing that isn&#x27;t a problem is indeed

Joe Lesueur · Answer

the following problem tells us that matrix A is m by n matrix Matrix B is a p by n matrix and we must prove that a times the transpose of be all transposed equals matrix b times the transpose of matrix A. To start this problem, we will use index notation. So a times that trans, whose of matrix b all transposed is I&#x27;m Jay Using that notation, we can take away the final transpose of the entire major cities, have a times transpose a matrix B and we can just foot those and put J i since we&#x27;re transposing the entire operation. And this is also equal to the summation cake. Want em of a J K be transpose Okay, I right now these case are just placeholders were mainly focusing on the the I N J notation. You can see a transfers to just the JK notation all the transpose of B to the K nine notation. I&#x27;ll find them together. The case with cancelling you&#x27;d have a B j I. So the next step of this problem would be to take away the transposed and move it to a We can set this equal to summation K equals 1 10 of be I today. We&#x27;re just taking that transpose here and applying it by switching these around and this is multiplied by a okay, J transposed. We&#x27;re doing the opposite here, or that gene K now flips because we&#x27;re adding the transpose operation. And now, since this notation can also be written back in the form of matrix be multiplied by the transpose of matrix

Griffin Johnston · Answer

suppose I had a square matrix. A So a is an in by in matric it has the same number of rows as column. And suppose also that the road space of a is equal to the null space of a no space of a And just as a quick reminder than no space of a is all the vectors x the set of all the vectors x such that a times the vector X is equal to zero. It&#x27;s all solutions to the homogeneous equation with this matrix. No, we want to prove in this video that a prove a the A cannot be convertible now, like I talked about in previous videos Ah, good first initial investigation for proof problems especially like this one. We&#x27;re trying to show that a or some some some mathematical object in this case AIDS and matrix. We&#x27;re trying to show that the A cannot have a certain property. In this case, we&#x27;re trying to show that they cannot be in vertebral, usually a good first initial approach to kind of get your hands dirty, and sometimes it ends up being the proof. It ends up turning into a foolproof, but regardless a good initial approaches to attempt to prove this by contradiction. So what is our goal? We try to prove something by contradiction. Well, we first suppose the contrary to suppose suppose a can be convertible can be convertible, then our goal, our goal when we suppose this sister derive a chain off implications. So if a driver change implications, what I mean by that is supposed taken, be in verbal without implies something. And that that&#x27;s something implies another something, and so on and so forth and to eventually we this the end of this chain of implications. We arrived at the end of his chain of implications. We arrive at some contradiction and maybe not the end, but one of these statements in here in this chain of contradiction, sometimes it&#x27;s we we stop when we arrive at a contradiction, so it ends up being the end of our chain. But it&#x27;s not necessarily the end of the Cheney. This infiltrated implications could go on forever, but regardless we eventually arrive at some contradiction. And when we arrive at a contradiction, that means that if we were supposed, if we were to suppose a can be in vertebral, then that would imply a contradiction. It would imply a statement that is absurd. That contradicts our original assumptions Are original assumptions that were given to us to be true for this particular problem. And thus this statement right here Supposing a can be in vertebral that can&#x27;t possibly drew a cannot be in vertebral if you were to arrive at a contradiction. Therefore, A Like I says, I said a cannot be convertible and we&#x27;ve proven a statement. So think for a minute what would happen if a is in verbal and make sure we want to leverage the fact that the road space of a in the north face of a are equal. So what would happen if a is and vertebral and we also have that the roast base of a is equal to the no space of a So I&#x27;m assuming you&#x27;ve had a go at it. So when a is in vertebral by the in vertebral matrix here, Um, all right, the in vertebral matrix here, um, there&#x27;s two things that we know. One of the things that we know if that if an matrix is in verbal, then the Onley solution x to the homogeneous equation A time of the vector X is equal to the zero vector. The only solution is the trivial solution. So X equals zero Vector is the Onley solution to this equation the Onley solution to this equation the Onley solution to this equation. But if X is equal to this year vector that that&#x27;s the only solution to this equation. Remember the set of all acts such that access solution to this equation is the knoll space of a So that means that the north face of a the north face of a is equal to the set containing zero vector and that&#x27;s it. That&#x27;s the only vector in the north place of a now something else that thean vertebral matrix theorem says is that if I&#x27;m matrixes in vertebral, which we are supposing a to b in vertical, remember by our by our our attempted a proof of contradiction proof by contradiction, we&#x27;re supposing that a is in vertebral. So if they were to be in vertebral then this will be true and also by the in vertical Matrix syndrome A is row equivalent to the end by in identity matrix And check out what we know we know that if two matrices are row equivalent, then that means they&#x27;re row spaces are equal. So the road space of a is equal, since a is row equivalent to the identity matrix because a convertible by the in vertebral matrix serum again, we&#x27;re supposing it to be in vertebral here, just just to see what happens to see if we arrive at a contradiction. So since we&#x27;re supposing it to be in vertebral, we know by the invariable major theme it&#x27;s row equivalent to the identity matrix. And we know there&#x27;s a theory that says that if two matrices aero equivalent, then that means that there rose spaces are the same. So we can say that the Roadway survey is the same as the rose base of the identity matrix, but the road space of the identity matrix, the in by an identity matrix. I&#x27;ll just write a kind of a scratch example of the identity matrix, so each vector looks a little bit like this, and finally we get to the last vector. In this I didn t matrix the identity matrix, the road space of the identity makers. It&#x27;s the span of all the row vectors now the span of all these row vectors is all of our in and namely this family&#x27;s row vectors for sure has the vector 1000 and so on, all the way to in or in minus one number of zeros. So the span of all these row vectors is definitely not equal to the set containing Onley, the viewer vector. Moreover, the span of all these row vectors actually equals all of our in So So let me let me emphasize this year. So So this span of all these reflectors is all in your combinations of these row vectors. So, for instance, one vector that would be in the row space off a would be the vector 100 all the way up two in minus one zeros bluff 01 all the way up two in minus one zeros, including this first year right here. And this vector is equal to 11 and then zeros for the rest of the entries. And this vector right here the fact that this factor is in the road space of a because any linear combination of the row vectors off the identity matrix, any linear combination of the row vectors. Identity matrix will lie in the span of the row vectors, a k a in the rose base off the identity matrix, which is the same as the space of a So therefore, the fact that this vector the linear combination of the other of the row vectors in this matrix, these two vectors plus zero times all the other vectors this vector lies in the road space of a. But remember, we suppose that the road space of a is equal to the no space of a. That means if the roads base of a sequel to the North based pay we were given this to be true. That means that every vector on the road space of Asia is in the north face of a and every vector in the north face of a is also in the rose face of a. But check this out. I found ah, vector in the road space of a That is definitely not in the north face of a. So the fact that a is in vertebral implies a contradiction, namely, the fact that we found a vector in the road space of a that is not in the north face of a implying that the road space of A is not equal to the North Bay survey which contradicts what we were given to be true. So if we were to suppose that a can be in vertebral, we would arrive at a contradiction at a contradiction. So therefore a cannot be in vertebral and we&#x27;re done.

Problem

Let $T$ be a one-to-one linear transformation fro…

Problem 13 Easy Difficulty

Let $A$ be any invertible $n \times n$ matrix. Show that for $\mathbf{u}, \mathbf{v}$ in $\mathbb{R}^{n},$ the formula $\langle\mathbf{u}, \mathbf{v}\rangle=(A \mathbf{u}) \cdot(A \mathbf{v})=(A \mathbf{u})^{T}(A \mathbf{v})$ defines an inner product on $\mathbb{R}^{n} .$

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David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Lily A.

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications