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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$\begingroup$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 $n$?
$\endgroup$ $\begingroup$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?
EDIT:
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.
$\endgroup$ 6 $\begingroup$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.
$\endgroup$ $\begingroup$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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