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Welcome back in this video. Im just going over how to find the null null space of a matrix so basically first off the notation for the null space. The matrix is the capital letter.

N. Followed by the name of the matrix. In brackets.

So. If our matrix. Is named a.

We write a capital n. With a in brackets in the null space of a matrix is the set of vectors that form the homogeneous solution that can be written like in this case. Where we have a we set it up as an augmented matrix and set the whole right hand side to zeros.

So the null space also happens to form a valid subspace thats called property of them. But basically what we want to do is if we want to solve or if we want to find the null space of matrix. A we have to set it up as a homogeneous system and this also represents the equation.

Where we have a times this vector x. Is equal to the zero vector or expand. It out its the same as this where we have a here and x.

Here. And then again this your l. Vector.

Just like that alright. So we can solve this augmented matrix by just finding the reduced row echelon form of the it so the first thing that we want to do is we apply our elementary row operations. And i think what we want to do is we want to do row.

3. Minus row. 2.

So lets lets first write in row 1. Were not going to be changing row. 1.

In here.

So we have 1 3. 2. In a zero row.

3. Minus. Row.

1. So were going to get 1 minus 1 is 0 1 minus 1 is 0 0 minus 0 is 0 0 minus 0 is 0 and then here for row two we can also just do it in this step. I think we want to eliminate this to be a 0.

So lets do row. 2. Plus row 1.

So we have 1 plus negative. 1. Is 0 1.

Plus. 3. Is 4 0.

Plus. 2. Is 2 and 0 plus.

0. Is 0 alright in the next step lets multiply row. One by negative 1 to switch this to a positive.

1 and then we will divide row. 2. By four because we want this leading entry to be a one as well so we have row 2 4.

And when we fill this out we get this becomes 1 this becomes negative 3 negative 2 and the 0 there and then here we have 4 divided by 4 is 1. 2 divided by 4 is 05 0. Divided by 4 is 0.

And we didnt do anything to this last row of zeros all right theres one more step that we can do here is we can get rid of this entry.

So well just have row 1 plus. 3. Times row.

2. Ok. So we get 1 plus.

3 times. 0. Is 1 negative 3.

Plus. 3. Times 1 is.

0. And negative. 2 plus 3.

Times 05 so thats negative. 2 plus. 15 that is negative 05.

All right the last one is still. 0 and then everything else is unaffected so we have positive 05. 0.

And then that last row of zeros all right so we have it in reduced row echelon form now and we can rewrite it back as a system of linear. Equations so we can write this as x. 1 minus.

05. Times x. 3.

Is equal to 0. Thats this first. Line.

Up.

Here and then the second line is x. 2. Plus.

05. X. 3.

Is also equal to 0. So what i like to do at this point is write what each variable is basically so we have x 1. Is equal to if we bring this over to the other side.

X. 1 is equal to 05. X.

3. And x. 2 is equal to if we bring this over it will be.

Negative 05. X. 3.

And then x. 3. Is just that in the pen and variable.

Its just equal to x 3. Its just chillin by itself. So now that we have all this we can look up here that then notice that the x vector was equal to have these components x1 x2 and x3 and we have the values here for them so we can rewrite this in vector form where we have zero point 5 x3 negative zero point 5 x3 and just x3 and because x3 can just really be anything just to simplify things we can actually just get rid of the subscripts now so if we just erase that basically were getting.

This vector that has this form of. 05. X.

Negative. 05. X.

And.

X. Where x. Can be absolutely anything.

And again. If thats the case. Then also what we can do is we can clean it up a little bit and just write this as because x can be anything well just multiply everything by 2 to get this to be 1 negative.

1. And 2. So these two vectors have the same form this one is just cleaner to look at and really we could rewrite any letter in here.

Theres naught to be x. We could have written a or something just to show you that if we pick a number here this one has to the second element has to be the negative of that and the third element has to be twice the the first element. So basically this vector here is the null space of a this is the set of all vectors.

With this form that are the solution to this homogeneous problem. So we can come up. Here.

And write. This. This is equal to x.

Negative. X. 2.

X. And. Then also if youre curious basically.

Like i said. This is a valid subspace and the basis of this subspace or the basis of the null space can just be written if you just take out the x basically. Its just 1 negative 1 2.

So all scaled up and scaled down versions of this basis form. The null space. Which is a subspace.

.

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