14. Let eij be matrix with one in the (i,j)th entry and zeros elsewhere Which 2 x 2 matrices eij can be added to the set below to form basis of R2,22 A-[9 #]-[? %]c...

Exercises $17-20$ refer to the matrices $A$ and $B$ below. Make appropriate calculations that justify your answers and mention an appropriate theorem.
$$
A=\left[\begin{array}{rrrr}{1} & {3} & {0} & {3} \\ {-1} & {-1} & {-1} & {1} \\ {0} & {-4} & {2} & {-8} \\ {2} & {0} & {3} & {-1}\end{array}\right] \quad B=\left[\begin{array}{rrrr}{1} & {3} & {-2} & {2} \\ {0} & {1} & {1} & {-5} \\ {1} & {2} & {-3} & {7} \\ {-2} & {-8} & {2} & {-1}\end{array}\right]
$$
Can each vector in $\mathbb{R}^{4}$ be written as a linear combination of the columns of the matrix $A$ above? Do the columns of $A$ span $\mathbb{R}^{4} ?$

In this video. We're considering the Matrix be further. So be was given to be this four by four matrix And what we've done already is we have read role reduced it into its echelon form. We found that row one, row two and row three in its role reduced form our pivot columns So we can say for the following that be in particular does not have a pivot in every row since the fourth row is a zeroed out row. Now this not having a pivot in every row causes certain problems. For example, we do also know as soon as there is not a pivot Every row that there exists vectors be in our four Where are four comes from? The four rows of this matrix p So there exists Exists Victor's being are for that are not linear combinations of the columns of be This is an important thing to consider because it means that in a vector equation such as B X equals, say why would not have a solution for every vector. Why? Let's consider a different situation. We know the set evolving your combinations is also the span of the columns of B and we also know that we have three pivots altogether. So let's consider the claim. The columns of be spans are three. This is an interesting claim because it really feels quite right. Sure, we don't have a pivot and every row, so the comes of B cannot spend our four. But if there are three pivots, could the calm spend our three? Well, Sadly, this claim is false. The columns of be do not span or three and this claim is false irrespective of the number of pivots. So if we had no idea what the number of pivots were, if we were just back to the original matrix, it would still be faults. And the reason is since B is in are four by four. In other words, this four here is the most important thing to consider and it causes a mismatch with our three. So the columns of beak on lee have a chance of spanning are for not are three, not are two and so on

In this example, we have a two by two matrix say that's provided medical. Here's to find its Aiken values and corresponding. I get vectors to start off for the first part. To find Eigen values, we take the determinant of a minus lamb die or will be solving for the very bold Lambda as a determinant will have negative Lambda Negative eight in the first column than one end for my is Lambda in the second column, Then, to take this to determine it will first have the main diagonal multiplied, which is negative. Lambda Times four Mice Lambda and subtract the product of the off diagonal, which is negative. Eight. So we'll have a positive eight so far. Then, if we simplify this, we obtain Lambda Squared minus four. Lambda Plus eight equals zero, and this is our characteristic equation. The solutions are Lambda One equals two to minus two. Lambda or excuse me to I and Lambda Two is that's constant to with a plus two I. So we confined that using the quadratic formula, and the next thing we're going to do is determine the corresponding Eigen vectors for each Eigen value. So let's focus on Lambda one To start out, we can solve the system a minus lambda one, which is two minus two I times X equals zero vector. So let's augment and say Our matrix is going to be distributing the negative sign in. In the first calm will have negative two plus two I and the negative eight below. In the second column, we have one. Then take four and at together the result of taking negative two plus two. I now we'll have altogether two plus two. I they were augmenting with a zero vector, which I'll place here. Now it's due to operations as we ro reduce. I want to go to the second row, divide it by negative eight, then make it become the first row will have one. Then negative 0.25 plus 0.25 I and a zero. Then the second row is going to be the old first row, which is negative. Two plus two I one and a zero. Now we have a pivot that's here and uncle is to eliminate this entry. So let's start by copying the first row. It is a one negative points to five plus 50.25 I and a zero. And what we'll be doing is multiplying Row one by two minus two i, and adding the results to the second row will obtain zero. And if we take negative points to five plus 50.25 I and multiply it by Tu minus two, I we obtain exactly negative one, and that combines with this entry producing a zero here. Then we have a zero for the last column where we augmented. So in the row one of our Matrix, which is in row reduced echelon form, we find that if we saw for the very bill, X one will be equal to the opposite of 0.25 plus 0.25 i times x two and X two is a free variable Salt. Just write X two equals X two Oh, it will have a negative here that was neglected to be copied from there. So now we need to describe what our first vector, Aiken vector V one, is that corresponds to Lambda One and we have some choices. If we decide to not deal with these decimal values we see here, we can allow X two to be any real number except zero since zero is never Nike in Vector. So is pick X two to be, say, a four. If we do that, then all together will get a one minus I. That's a result of placing for here distributing as well as distributing the negative sign. Now we have to re piece repeat this row reduction to find the second I get vector, but there is a wonderful shortcut. That shortcut says that V two is the congregate of the one and the one conjugated would be one plus I and the contra gave a real number is itself. So what we've done is we found dykan values V one, V two and the corresponding Eigen vectors V one and V two. So these V one V two also form a basis for their corresponding Eigen spaces.

If we were given the following set of vectors and if you were asked to find a basis for the span of this set of actors in one way, we could find a basis for the span of the set of vectors is to make this vector the first row of the Matrix, this vector, the second rover matrix in this vector, the third row of the major. That's to say, make the rows of the matrix correspond to each of these vectors. So let me show you what I mean by that. So what we did here is we made a matrix whose rose corresponds exactly to the vectors in this set right here. So check this out. Based off of the definition off aerospace, off a matrix, The roads base of the Matrix is the span of its row vectors. So since the row vectors in this matrix correspond exactly to the vectors in the set of vectors, if we find the basis for the roast base of his matrix, we will have also found ah, basis for the span of the set of victories. Because again, the row space is a span of its row vectors. So finding a basis for the aerospace that this matrix is the same spine. Ah, basis for the span of the set of vectors. So to find a basis for the most base of this matrix, we first we first row, reduce ro reduce matrix to a row echelon for and then next we take the non zeros take the non zero both in those lows. Those non zeroes off the row echelon form matrix form a basis for the road space of this matrix. And the reason why this works, I discussed in previous videos where we talk about finding the basis for the road space of the Matrix. So if you're curious about why that works, go ahead and check those out. So take a minute. Impossibly dio and load is this metric toe a row, ash on form. I'm assuming you've had a go at it. I'm gonna reveal the Siris of low equivalent sees that leads us to a Rochon form of this matrix. So these low equivalence ease leave this to or this Roque, the Siris of Roe equivalent sees, leads us to this row echelon for matrix right here in the non zero rows of this role echelon for matrix. This room right here and this one over here Thies to row vectors form a basis for the road space of this matrix, which is the roadways of this matrix is the same as a span off this set of vectors. So a basis for this set of the span of the set of actors is the set of exes containing the vector one comma, one common negative, one comma too. The first non zero grow in this revolution for matrix and a vector zero comma Negative one comma five common negative eight. No, we could do the same thing, but using the notion of a column space. So if we made this vector rather, these vectors in this set correspond exactly to the column vectors and in Matrix. So let me show you what that means. So the column vectors in this matrix correspond Exactly. You are exactly the vectors in this set of vectors. So that means if we find ah basis for the column space of this matrix, the Mueller also found a basis for the span off the set of vectors because the column space of this matrix is the span of the car vectors and the column vectors in this matrix are the same as the vectors in this set of vectors. So take a minute. Posit Video Road Is this major or rather? Let me show you that. Let me first review the method off finding a basis for the class based on a matrix. So we first wrote, Reduce matrix to a row echelon form a row echelon form. We'll call that Rochelle on full on P Prime and let's call this Matrix P, as they usually do so, produce major toe Aurora Sean form P Prime to find the pivot columns to find the pivot columns of P. Because the pivot columns of P Prime correspond exactly to the pivot columns of P, Let's say I found out that the pivot cons of P Prime are the first and third columns of P prime. Well, then, that means that the first and third columns of P are the pivot columns of P. So once we find those pivot cons, we take the pivot columns of P of the original Matrix B. This matrix right here, this matrix right here we take the pivot cons of peace or take pivot columns of P and those columns form a basis for the calm space of P. So all lets you find a row echelon form of this matrix p to go ahead and possibly dio. So I'm assuming you've had a go at it. So I'm going to reveal the Siris of Roe. Equivalency is that leads us to a row echelon form of this matrix p. So this is the series of Roe equivalent sees that leads us to a row echelon form of the matrix P of this matrix pee right here. And this is our role echelon form. And this role echelon form reveals where the pivot cons appear located. Because this column in this calm or a pivot com off this relation gonna call this relish on for matrix p prime for this column in this column Occam's with leading entries comes one and comes to are the pivot comes a p prime. So therefore, the first and second columns of P. R. The pivot calms API, and we know that the pivot columns of P former basis for the calm space of P and the common base of P is the same as the span of the set of vectors. So we found a basis for the span of the set of X right here, namely the set of vectors containing the vector one comma, one common negative, one comma too. And the vector to come One comma, three common negative four in the set of vectors right here is a basis for the calm face of this matrix, which is the same as a span of the set of actors. Now, you might be wondering, Why does this fan of the set of vectors have two different bases? But remember, So first off, first off, this span of the set of vectors is a subspace of our four. Why is it a subspace of our four? Because the span of the set of vectors any vector in the span of the set of vectors is a linear combination off these vectors and Angelina linear combination of these vectors will contain four rial number components. That is, it is a vector in our for the set of all vectors containing four real number components. So remember, a subspace can have multiple bases When he said it again. A subspace can have multiple bases and Here's a good example of a subspace having at least two bases. They can have more. It can have more basics. It doesn't have to have one basis. So here are two bases for one vector, one subspace of our four, namely the span of the center vectors right here.