The rank of an $m \times n$ matrix $M$ is the dimension of the range $\calR(M)$ of the matrix $M$.
The range of the matrix $M$ is
\[ \calR(M)=\{\mathbf{y} \in \R^m \mid \mathbf{y}=M\mathbf{x} \text{ for some } \mathbf{x} \in \R^n\}.\]

Proof.

(a) $\rk(AB) \leq \rk(A)$.

Recall that the rank of a matrix $M$ is the dimension of the range $\calR(M)$ of the matrix $M$.
So we have
\[\rk(AB)=\dim(\calR(AB)), \quad \rk(A)=\dim(\calR(A)).\]

In general, if a vector space $V$ is a subset of a vector space $W$, then we have
\[\dim(V) \leq \dim(W).\]
Thus, it suffices to show that the vector space $\calR(AB)$ is a subset of the vector space $\calR(A)$.

Consider any vector $\mathbf{y} \in \calR(AB)$. Then there exists a vector $\mathbf{x}\in R^{l}$ such that $\mathbf{y}=(AB)\mathbf{x}$ by the definition of the range.
Let $\mathbf{z}=B\mathbf{x} \in \R^n$.

Then we have
\[\mathbf{y}=A(B\mathbf{x})=A\mathbf{z}\]
and thus the vector $\mathbf{y}$ is in $\calR(A)$. Thus $\calR(AB)$ is a subset of $\calR(A)$ and we have
\[\rk(AB)=\dim(\calR(AB)) \leq \dim(\calR(A))=\rk(A)\]
as required.

(b) If the matrix $B$ is nonsingular, then $\rk(AB)=\rk(A)$.

Since the matrix $B$ is nonsingular, it is invertible. Thus the inverse matrix $B^{-1}$ exists. We apply part (a) with the matrices $AB$ and $B^{-1}$, instead of $A$ and $B$. Then we have
\[\rk((AB)B^{-1}) \leq \rk(AB)\]
from (a).

Combining this with the result of (a), we have
\[\rk(A)=\rk((AB)B^{-1}) \leq \rk(AB) \leq \rk(A).\]
Therefore all the inequalities are in fact equalities, and hence we have
\[\rk(AB)=\rk(A).\]

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The dimension of the nullspace of $A$ is called the nullity of $A$.
Prove the followings.
(a) $\calN(A)=\calN(A^{\trans}A)$.
(b) $\rk(A)=\rk(A^{\trans}A)$.
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Hint.
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Let $A$ and $B$ be $m\times n$ matrices.
Prove that
\[\rk(A+B) \leq \rk(A)+\rk(B).\]
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(a) Let $A=\begin{bmatrix}
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1 & 3 & 1 & 2
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Find a basis for the range $\calR(A)$ of $A$ that consists of columns of $A$.
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Let $V$ be the following subspace of the $4$-dimensional vector space $\R^4$.
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x_1 \\
x_2 \\
x_3 \\
x_4
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x_1-x_2+x_3-x_4=0 \quad\right\}.\]
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3& 2
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and consider the following subset $V$ of the 2-dimensional vector space $\R^2$.
\[V=\{\mathbf{x}\in \R^2 \mid A\mathbf{x}=5\mathbf{x}\}.\]
(a) Prove that the subset $V$ is a subspace of $\R^2$.
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Determine whether the following is true or false. If it is true, then give a proof. If it is false, then give a counterexample.
Let $W_1$ and $W_2$ be subspaces of the vector space $\R^n$.
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Let $A$ be an $n\times n$ idempotent matrix, that is, $A^2=A$. Then prove that $A$ is diagonalizable.
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