Big Picture of Linear Algebra

Let \(A\) be a \(m\times n\) matrix.

Theorem: The dimension of the column space of \(A\) is \(r=\text{rank }A\).

Proof: \(\text{rank }A\) is the number of linearly independent columns in \(A\). That is, the dimension of the column space of \(A\).

Theorem: The dimension of the row space of \(A\) is the same as the dimension of the column space of \(A\).

Proof: \(\text{rank }A^t=\text{rank }A\).

Proof: Elementary row operations DO NOT change the row space of \(A\). Elementary row operations MIGHT change the column space of \(A\). However, elementary row operations do not change the linear dependence of the columns of \(A\). Consider the reduced row echolon form of \(A\). The number of linearly independent columns of \(A\) is the same as the number of linearly independent rows of \(A\).

Theorem: The dimension of the null space of \(A\) is \(n-r\).

Proof: This follow immediately from the Dimension Theorem.

The right part of the picture can be proved by considering \(A^t\) and follows immedicately from the left part.

Theorem: Denote the row space of \(A\) as \(V\) and the null space of \(A\) as \(W\). We prove that \(\mathbb{R}^n=V\oplus W\)

Proof: By the Dimension Theorem, we have already known that \(\dim{\mathbb{R}^n}=n=\text{rank }A+\text{nullity }A=\dim{V}+\dim{W}\).

We show that \(V\cap W=\{\vec{0}\}\). Suppose that \(u\in V\cap W\). Since \(V\) and \(W\) are orthogonal, we have \(\langle u, u\rangle=0\). This follows that \(u=\vec{0}\).

We can choose a basis \(\alpha=\{v_1, v_2, ..., v_{\dim{V}}\}\) for \(V\) and a basis \(\beta=\{w_1, w_2, ..., w_{\dim{W}}\}\) for \(W\). Then \(\gamma=\{v_1, v_2, ..., v_{\dim{V}}, w_1, w_2, ..., w_{\dim{W}}\}\) is linearly independent because \(V\cap W=\{\vec{0}\}\). Combining this and \(\dim{\mathbb{R}^n}=\dim{V}+\dim{W}\). We have that \(\gamma\) is a basis for \(\mathbb{R}^n\).