Problem 18 Hard Difficulty

Let $A$ be an $n \times n$ matrix with rowspace( $A$ ) $=$ nullspace( $A$ ). Prove that $A$ cannot be invertible.

Answer

Hint: Try to prove by contradiction and use the Invertible Matrix Thereom.

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Calculus 3

Differential Equations and Linear Algebra

Chapter 4

Vector Spaces

Section 8

Row Space and Column Space

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Video Transcript

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'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're trying to show that a or some some some mathematical object in this case AIDS and matrix. We're trying to show that the A cannot have a certain property. In this case, we'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'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's we we stop when we arrive at a contradiction, so it ends up being the end of our chain. But it'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'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'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'm assuming you'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'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'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's it. That's the only vector in the north place of a now something else that thean vertebral matrix theorem says is that if I'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'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'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'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're supposing it to be in vertebral, we know by the invariable major theme it's row equivalent to the identity matrix. And we know there'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'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'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'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're done.

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