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4.D Dimension theorem

In understanding these subspaces, one of the first questions that might come to mind is: “How big are they?” Well, the size of a subspace is measured by its dimension, and the following theorem shows that if you know the dimension of either $\operatorname{im}\nolimits T$ or $\ker T$ , then you immediately also know the dimension of the other one.

Theorem 4.24 (Dimension theorem).

Let $T:V\to W$ be a linear transformation between vector spaces over $F$ , where $V$ is finite dimensional. Then

\dim(\operatorname{im}\nolimits T)+\dim(\ker T)=\dim V.

[Aside: This is sometimes also called the “Rank-Nullity theorem” because $\dim(\operatorname{im}\nolimits T)$ is the rank of $T$ (see below), and $\dim(\ker T)$ is often referred to as the nullity of $T$ .]

Proof.

Let $n=\dim V$ , and choose a basis $(\vec{x_{1}},\cdots,\vec{x_{k}})$ of the kernel, where $k=\dim(\ker T)$ . By Corollary 2.37, we can extend this linearly independent set to a basis of $V$ , by adding vectors $(\vec{x_{k+1}},\cdots,\vec{x_{n}})$ . Now I claim that $\mathcal{B}=(T(\vec{x_{k+1}}),\cdots,T(\vec{x_{n}}))$ is a basis for $\operatorname{im}\nolimits T$ .

Since the image of $T$ is spanned by the images of the basis vectors, and $T(\vec{x_{i}})=\vec{0}$ for any $i=1,\cdots,k$ , this shows $\mathcal{B}$ spans $\operatorname{im}\nolimits T$ .

To show $\mathcal{B}$ is linearly independent, assume we have scalars $\alpha_{i}\in F$ such that

\vec{0}=\sum_{i=k+1}^{n}\alpha_{i}T(\vec{x_{i}})=T(\sum_{i=k+1}^{n}\alpha_{i}% \vec{x_{i}})

, where last equality follows from the linearity of $T$ . In particular, this means $\sum_{i=k+1}^{n}\alpha_{i}\vec{x_{i}}\in\ker T=\operatorname{span}\{\vec{x_{1}% },\cdots,\vec{x_{k}}\}.$ So by the linear independence of the $\vec{x_{i}}$ , we have $\alpha_{i}=0$ for all $i$ . This proves $\mathcal{B}$ is linearly independent, and hence is a basis for $\operatorname{im}\nolimits T$ . Therefore, $\dim(\operatorname{im}\nolimits T)=n-k=\dim V-\dim(\ker T)$ as required. ∎

We will define the rank of a matrix differently to that used in MATH105. The “rank” of $A$ is defined as

\operatorname{rank}A:=\dim(\operatorname{im}\nolimits A).

The next theorem shows that this definition is equivalent to the one used in MATH105.

Theorem 4.25.

Let $A\in\operatorname{M}_{{n\times m}}({F})$ be a matrix. Then

\operatorname{rank}A=\dim{\textup{(span of columns of $A$)}}=\dim{\textup{(% span of rows of $A$})}.

Proof.

The left equality is true by Theorem 4.16. The proof of the right hand equality is omitted from this module, but we include it below for the interested reader.

Let $A_{red}$ be the reduced row echelon form of $A$ . By Theorem 4.24, we have

\operatorname{rank}A+\dim(\ker A)=\operatorname{rank}A_{red}+\dim(\ker A_{red}).

But $\dim(\ker A)=\dim(\ker A_{red})$ by Theorem 4.23, and hence $\operatorname{rank}A=\operatorname{rank}A_{red}$ .

Next, by Theorem 2.49, the number $r:=\dim{\textup{(span of rows of $A_{red}$)}}$ equals the number of non-zero rows of $A_{red}$ , and therefore the image of $A_{red}$ is contained in the $r$ -dimensional subspace $\operatorname{span}\{\vec{e_{1}},\cdots,\vec{e_{r}}\}\subset F^{n}$ . So $\dim(\operatorname{im}\nolimits A_{red})\leq r$ . In other words:

\dim{(\textup{span of columns of $A$})}=\operatorname{rank}A=\dim(% \operatorname{im}\nolimits A_{red})\leq r=\dim{(\textup{span of rows of $A$})}.

Since this argument applies to any matrix, it applies to the transpose $A^{T}$ , which tells us the reverse inequality is true (since the transpose operation exchanges the rows and columns of a matrix). Hence we must have equality. ∎

An immediate consequence of this Theorem is that $\operatorname{rank}A=\operatorname{rank}A^{T}$ .

Exercise 4.26:

Let $D:\mathcal{P}_{3}({\mathbb{R}})\to\mathcal{P}_{3}({\mathbb{R}})$ be the linear transformation defined by differentiation of the single variable. For example, $D(x^{2})=2x$ . Let $\mathcal{B}=(1,x,x^{2},x^{3})$ ; this is the standard basis for $\mathcal{P}_{3}({\mathbb{R}})$ .

i.

Compute $[D]_{\mathcal{B}}$ ,
ii.

Find a basis for the kernel of $D$ ,
iii.

Find a basis for the image of $D$ ,
iv.

Verify the dimension theorem for $D$ .

[End of Exercise]

Corollary 4.27.

Let $A$ be a square matrix. Then the following three conditions are equivalent to each other:

•

The rows of $A$ are linearly independent
•

The columns of $A$ are linearly independent
•

$A$ is invertible.

The above corollary to Theorem 4.25 is used regularly in statistics during the process of multiple linear regression (see MATH235 and MATH452).