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If $A$ is any $5 \times 5$ invertible matrix, then what is its column space? Why?

I'm totally lost with column space. Any ideas?

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Hint: given a vector, how do you whether it belongs to the column space or not? – Kirill Jun 25 '13 at 1:48
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If $A$ is invertable, is there any vector $v \in \mathbb R^5$ which cannot be composed by a linear combination of the columns of $A$ (i.e. the column space)? What does that say about the column space?


I can't imagine that I am providing anything that you haven't read before, but here goes...

Span is the set of all possible linear combinations of a set of vectors. In the simplest case the span of a single vector $v$ is a line. If we have two vectors, the span could be a line or a plane; if they are linearly independent, then they span a plane. The simplest example of this is the familiar coordinate plane spanned by $\vec i$ and $\vec j$. Any point can be represented by $a\vec i+ b\vec j$ for some $a$ and $b$. In that case we say that $\vec i$ and $\vec j$ span the plane and because they are linearly independent they also form a basis for a 2-D space (i.e. $\mathbb R^2$). Just for kicks, let's represent the coordinate point (2,0); it is clear that $2\vec i+ 0\vec j$ represents that point.

We like $\vec i$ and $\vec j$ because they are easy to work with but the idea of span extends to any set of vectors. For example, we could have just as easily chosen $\vec v = \vec i + \vec j$ and $\vec u = \vec i - \vec j$ and because they are linearly independent we can also represent any point in the coordinate plane as a linear combination represented as $c\vec v+ d\vec u$. Now let's revisit our coordinate point (2,0); can you see how $1\vec v+ 1\vec u$ represents the same point?

We can continue the same thought process to 3, 4 and 5 dimensions and so on...

For your question we have five vectors arranged in a matrix $A$. Each vector is taken as a column of the matrix. If we have a vector represented as a $5\times 1$ column vector $x$ then the product $Ax$ represents a linear combination of the columns of $A$ by the weights in $x$. Thus for the equation

$$Ax = b$$

the linear independence of the columns of $A$ tell us that we can represent any point $b \in \mathbb R^5$ as a linear combination of the columns of $A$.

So the bottom line is that in order to solve $Ax = b$, the vector $b$ has to be in the column space of $A$. That is why column space is important.

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I have no idea. I'm totally lost on this whole concept. I've watched hours of videos on vector space and column space, read the chapter 12 times, looked over my notes, and am still lost. – Tyler Murphy Jun 25 '13 at 2:38
Where are you in your understanding? Do you understand linear combinations, span, dimension and linear independence? Do you understand the concept of a space? – Tpofofn Jun 25 '13 at 3:23
I'm good with linear combinations, kind of understand a span but not terribly comfortable with it, solid on dimension and linear independence. – Tyler Murphy Jun 25 '13 at 3:25
is the only way we have of knowing if b is in the column space is to see it intuitively or to solve the matrix by elimination? – Tyler Murphy Jun 25 '13 at 5:34
You have to take the augmented matrix to row reduced echelon form. Except for very simple cases you cannot determine it intuitively. – Tpofofn Jun 25 '13 at 11:46

This problem is a test of your knowledge of some important basic results about invertibility and rank.

If an $n\times n$ matrix is invertible, its rank is $n$, and vice versa. The dimension of the column space of the matrix is its column rank, and the dimension of the row space is its row rank. A basic result is that the two are equal, and we just speak of the rank of the matrix. Thus, if an $n\times n$ matrix is invertible, its column space has dimension $n$. The column space is a subspace of $\Bbb R^n$. What is the only subspace of $\Bbb R^n$ of dimension $n$?

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It's the whole space. An invertible matrix has kernel $\mathbb \{0\}$ and range $\mathbb R^5$ (or $\mathbb C^5$ or the $5$-fold product of something else, depending on what field you're working over). Recall the range is equal to the column space.

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An $m \times n$ matrix $A$ determines a linear map $L_A :\mathbb{R}^n \to \mathbb{R}^m$, defined by $L_A(x) = Ax$. Now if $A$ is an invertible $n \times n$ matrix, $L_A$ has an inverse. Hence, $L_A$ is bijective.

Knowing this, what is the column space of $A$? That is, what is the image of $L_A$?

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