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Quick example calculating the column space of a matrix. And the null space of a a matrix. We will first start by determining the column space of this matrix right by definition.

The column space of a is the span of all the columns of a which by definition is also the set of all linear combinations of the columns of a thus we can easily find the column space the definition of the column space is that is is the span of all the columns of a so i just take all my columns. All my columns and then i take the span of those columns. Which again by definition is the set of all linear combinations.

So its a set of all linear combinations of our vectors. But we can simplify this answer if i take a look at this answer. I am.

Using four vectors. Ive got these four vectors right here describing a subspace of r3 so all of these vectors. You know all of these vectors here they live in r3 thus.

When i take the span of vectors that live in r3 if i take the linear combination of vectors that live in r3. Im going to get more vectors that live in r3 so im looking at a subspace of r3 since im looking at all the nir combinations of a few vectors in r3.

We do not need four vectors to describe a subspace of r3. So we know we can simplify our answer. So we need to simplify that answer do you turman the fewest number of vectors.

We need to describe our column space. We can put a into epsilon form once its in escalon form. We can determine our pivot columns.

So youve got our pivots right here our first non zero terms give us our pivots and once we have our pivots. We can get our pivot column. So weve got these three are our pivot columns.

Well those pivot columns also correspond to pivot columns in our original matrix now that we know the pivot columns. We can just take the pivot columns from our original matrix. So the pivot columns from our original matrix.

And that will give us a basis for our column space. So our basis from the column space consists of our three pivot columns from our original matrix from the original matrix.

Notice that our basis for this particular example contains exactly three vectors. Ive got one two three elements in my basis. Because we are looking at subspaces of our three our spaces will always be finite dimensional.

So we just have a finite number of elements in our basis. So exactly three elements and since we have exactly three elements that means our column space will be three dimensional and our column space remember consists of all linear combinations of our three basis vectors and so there are an infinite number of elements in calais. We have all linear combinations.

So i can have one times this vector plus pi times. This vector minus three times that vector so that would be in call a any linear combination. I can choose any real numbers for my c sub.

Is so we have an infinite number of elements in our column space and since our column space is a set of all linear combinations. That by definition is the span. And so.

Our answer here is that our column space here is just the span of our first three columns can you identify call a so since we know that calm our particulars particular example. Our column space is three dimensional.

Because we have three basis elements. You should be able to identify call a i will leave that for a leave that answer for a later video. But hopefully you can answer since this thing is a three dimensional subspace of r3.

There is only one three dimensional subspace of r3 so you should be able to identify call a we will now determine the null space of the same matrix a to determine the null space. We not only need excellent form. But we need reduced echelon form.

The reason we need reduced echelon form is that by definition. The null space of a is exactly the solution set of our ax equals to 0. So that means we need to solve our ax equal to 0 in order to find the solution set because that is by definition.

The null space of our matrix a so once we have it in reduced echelon form. We can easily solve this one. Weve done that in a previous video.

So we can see that x. 1.

Is. Minus 2 x. 4.

Etc. And we now have our solution. Set.

So. This is our solution set right here to our ax equals to 0. And that means.

It is also our null space. So our solution set it consists of all multiples of this vector right here since it consists of all multiples of a vector in other words. Its all linear combinations of this single vector here it is the span of this and so our null space is the span of this one vector.

So it is all linear combinations. Or multiples. And so we have a one dimensional null space.

So this is the null space of our matrix a .

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