suppose-a-is-an-n-times-n-matrix-with-the-property-that-the-equation-a-mathbfxmathbf0-has-only-the-t

Question

Suppose $A$ is an $n 	imes n$ matrix with the property that the equation $A \mathbf{x}=\mathbf{0}$ has only the trivial solution. Without using the Invertible Matrix Theorem, explain directly why the equation $A \mathbf{x}=\mathbf{b}$ must have a solution for each $\mathbf{b}$ in $\mathbb{R}^{n}$ .

Runpeng Li · Accepted Answer

problem. Um, we want to show that if a X equals zero has only the trivial solution, she will solution. Then way. Want to show Dad? Hey, X, Because be must I have this ocean for every Be okay. Now, The way to do it without the convertible matrix serum is to observe the structure off A. So we consider let&#x27;s consider the reduced row ish in form off, eh? Say a prime. Okay, so scuse me. So that means, um, a prime times X equals zero. This system has has no pre variable because this system has has only has only intriguing solution that it means the system has no free bearable. Because we we know from our previous material we know that, uh, the system has nontrivial solution. It has at least a work repairable. So he&#x27;s just a negation of the statement. So has I&#x27;ll just write down here just for your information, A prime axe times. They&#x27;re a prime axe equal zero has has on Lee Trivial solution. It&#x27;s equivalent to say that this system has no free bearable. No, for you bearable. Okay, So for the Moore Demi&#x27;s matrix A has, um, has an people in positions. Of course. We&#x27;re assuming Matrix a had Matrix A&#x27;s On By and Square matrix. Yeah, this is our assumption. Um, people positions. Excuse me, people. Positions. And this is our conclusion here. Now, from here, we can say, um, since there are n people, people in positions and the matrix structure off, eh, Um, or we consider a prime and B and be seen as some some some values on the diagonal and zero for the lower triangular. An appetite for dinner. We have some numbers. I don&#x27;t. I just, uh, using darts to represent them. So, hey, in terms off a a prime time zags, we can see that of all the terrible here we are using a, uh, using this form of a prime then our nest our last variable se X end must be determined by the value we&#x27;re taking for B. So that means So we&#x27;re saying be one. So let&#x27;s suppose this is our vector B until fiend is equal to a prime times axe. And this is some sort of early on the diagonal times the vector X. So that means we were supposed defector ads his ex Juan x two txn Okay, So that means the last the last term Other day that I&#x27;ll be there diagonal If we multiply, multiply this This value by the nest. Last entry off. This factor acts. We have some. Aye, aye aye. Which is the entry for the last in the center of the diagonal of ape. Right? So? So there&#x27;s a II times the last the last available off the vector acts, which is X end is exactly the end. So we can solve for accent here. So that is bien divided by a II and so on so forth. So on so forth we can solve for everything Go bearable x one Until we were done, we have done for X one because, um, there&#x27;s no free bearable. So that&#x27;s why we can do this. So that means xn. That means acts the vector must exist. So what that means for the more damages the system a axe? It was B has a solution or every single b we&#x27;re taking the knob should be from are in

Michael Jacobsen · Answer

in this example, we&#x27;re starting off with a square matrix, which is going to be of size and buy in. Next. We have a matrix equation, A X equals B, and we&#x27;re told that we have at least one solution, and this is for every vector. B from RN. Let&#x27;s see what consequences come from these a statements. First, we know that this information implies that the Matrix A has a pivot in every row that comes from the statement that a X equals B is effectively always consistent for every single vector be coming from RN. That only happens if and only if there&#x27;s a pivot in every single row of our matrix. A. Now here&#x27;s where it being a square matrix is really important. If there is a pivot in every row and it&#x27;s of size and buy in, then this implies A has a pivot in every column as well. So a not being a square matrix gets a stuck on the statement and blue, but it being square allows us to go forward. So now that we know that there is a pivot in every column, let&#x27;s reconsider this matrix equation. I will say. Therefore, a X equals B has no free variables. The reason this statement follows from the last is because as soon as we have a pivot in every column, there can never be a free variable since a free variable corresponds to a non pivot column. So now that we know that a X equals B has no free variables and we also know that it is always consistent, this further implies that a X equals B has a unique solution. And this is for all vectors, be that come from our end. So this is an interesting situation. We&#x27;re told, if a X equals B has at least one solution and this is for every being our end, we conclude immediately that that solution is unique. But be very careful. This works on Lee when the Matrix is a square matrix.

Angelo Rendina · Answer

say we have the system A X equals B and disease Stoma solutions for every vector be in RN Now, if a he&#x27;s not in veritable, then we reduce the system metrics, eh? Me to something like upper triangle medics here and then on entire roll off zero at the end and appear this whatever. And then here we have the transformer coefficients so easy A is not in veritable. But that means that for some be hence be teal that the system cannot have solution against assumption that the system had solutions for every vector B. And of course, the problem is that we&#x27;re think we&#x27;re assuming that B is not in veritable so for the system to have solution always, the medics say, has to be in vertebral.

Angelo Rendina · Answer

we have the equation A minus. Hey, X Everything Immers equal Ex simmers me Now the first type is to show that bee is in veritable And to do so we multiply from the left by a minus I x so on the left hand side we&#x27;re left with identity and on the right Inside we have a minus a x ex embers Me But now the expression here in blue is by definition the inverse off me and so bees in vegetable. Okay, now that we know this, we can That&#x27;s right. Equation a minus x inverse equal X embers be by inverting both sides to obtain a minor sex equals being birth X. Remember that we invert the product. We get the product off the others in the inverse order. Okay, Now we can move the exxon aside. So let&#x27;s say we write a He&#x27;s equal to a plus being birth X. And now if we show that these metrics in brackets is in vegetable, that we can just multiply by theme verse to have axe equal to something and we&#x27;re done So to do that, we observe here that since a is in vertebral, we can multiply by eating birds from the right to obtain identity is equal to a Los beavers X a m birth. Now these means that X A chambers is the inverse Oh, a plus B in verse which is therefore is in vertebral. So now that we know that we can multiply from the left by the inverse off a plus by beavers And so we see a plus B in verse everything members, they&#x27;re multiplies, eh is equal to X and we have solved for X.

Viswesh Uppalapati · Answer

So, uh, in this in this video, we&#x27;re gonna be solving problem. Number 40 from the book from Section 1.7 is on linear Independence. So it gives us a statement and tells us to explain why it&#x27;s true. So it tells us that there&#x27;s an M by N matrix a wood, um, that has n pivot columns. So this means that and must be equal to M or must be greater than equal then because, uh, if there&#x27;s more if N is greater than our greater than M, then that means we cannot have end to the columns because there is. There aren&#x27;t enough Rose and the Matrix for, um all of every end toa have its own pivot. So where these Criterion line it tells us that, uh, the vector equation X equals B for this particular n by N matrix, which is a has must have a unique solution for every B. Um, this is true because, uh, if you set sent this equation equals zero X equals zero. This, uh, this problem or this n by N matrix will always have a trivial solution. If it has, Ah, if it has a unique solution for every B because being linear independence basic means that any vector that is made out of a linear combination of the columns of A has a unique representation, which means that there&#x27;s only one X so one solution for every, uh, X equals for every B and the equation X equals B. In other words, if we if you were to write this out, that means that, um, if and has a pivot every column, that means that, um so that means that there will always there there will always be ah, no free variable, so there will never be a free variable. So, for example, if you take, let&#x27;s say, uh, four by three matrix with a ah, give it and every every column because there are three columns. But for Rose, so the last row can just be zeros, and these can be a non zero numbers. We can see that if we label each column with a variable cause, um, because solving an augmented matrix is just solving a set of system equations, Um, where each column represents a different variable. You know that every call him having a pivot makes it so that there is no free variable. Um, which means that which means that every variable has one single unique solution. That is why this statement is true.

Michael Jacobsen · Answer

in this video, we&#x27;re starting off by assuming that the Matrix equation a X equals B will have at most one solution for all vectors, be that we might choose. Let&#x27;s take some logical steps to see where we can get from this statement. Well, if this works where we have at most one solution for all be, let&#x27;s let be be equal to the zero vector. It should work as well. For the zero vector by the statement, then a X equals zero vector has at most or at most is probably the most important key word here. One solution. So we know X equals zero vector is going to be consistent and we&#x27;re only going to get to one solution. But we already know that the Vector X equals zero vector is the trivial solution to this equation. So because we have a homogeneous equation with those euro vector here, placing as euro vector in this position gives us a solution which we call the trivial solution. But that means we have found one solution and there is only one solution because of the word at most. So let&#x27;s say what that means Next were right to therefore a X equals zero vector has on Lee the trivial solution thanks to the at most statement here again, probably the most important statement in this entire problem. No, when we say X equals of zero Vector has on Lee the trivial solution that gives us another conclusion immediately about this matrix A. We have now that the columns of a are linearly independent. So we&#x27;re able to say a very strong statement about the columns of a really, just from this information, which is quite surprising.

Problem

In Exercises 33 and $34, T$ is a linear transform…

Problem 32 Hard Difficulty

Suppose $A$ is an $n \times n$ matrix with the property that the equation $A \mathbf{x}=\mathbf{0}$ has only the trivial solution. Without using the Invertible Matrix Theorem, explain directly why the equation $A \mathbf{x}=\mathbf{b}$ must have a solution for each $\mathbf{b}$ in $\mathbb{R}^{n}$ .

Answer

Hence, the equation $A \mathbf{x}=\mathbf{b}$ must have a solution for each b in $\mathbb{R}^{n}$ .

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

Linear Algebra and Its Applications

Grace H.

Catherine R.

Maria G.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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Hence, the equation $A \mathbf{x}=\mathbf{b}$ must have a solution for each b in $\mathbb{R}^{n}$ .

Linear Algebra and Its Applications

Grace H.

Catherine R.

Maria G.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Catherine R.

Maria G.

Kristen K.

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

Linear Algebra and Its Applications