Fix the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$, and let $\R^3$ be the vector space of $3 \times 1$ column vectors. Define
\[W = \{ \mathbf{v} \in \R^3 \mid \mathbf{b} \mathbf{v} = 0 \}.\] Prove that $W$ is a vector subspace of $\R^3$.

Proof.

We verify the subspace criteria: the zero vector $\mathbf{0}$ of $\R^3$ is in $W$, and $W$ is closed under addition and scalar multiplication.

First, the zero element in $\R^3$ is $\mathbf{0}$, the $3 \times 1$ column vector whose entries are all $0$. Then clearly $\mathbf{b} \mathbf{0} = 0$, and so $\mathbf{0} \in W$.

Next, suppose $\mathbf{v} , \mathbf{w} \in W$, and $c \in \mathbb{R}$. Then $\mathbf{b} \mathbf{v} = \mathbf{b} \mathbf{w} = 0$, and so
\[\mathbf{b} ( \mathbf{v} + \mathbf{w} ) = \mathbf{b} \mathbf{v} + \mathbf{b} \mathbf{w} = 0.\] Thus, $\mathbf{v} + \mathbf{w} \in W$.

Because, again, $\mathbf{b} \mathbf{v} = \mathbf{0}$, we have
\[\mathbf{b} ( c \mathbf{v} ) = c \mathbf{b} \mathbf{v} = c \mathbf{0} = \mathbf{0}.\] Thus $c \mathbf{v} \in W$. These three criteria show that $W$ is a vector subspace of $\R^3$.

Comment.

We can generalize the problem with an arbitrary $1\times 3$ row vector $\mathbf{b}$.

The proof is almost identical.
(Look at the proof. We didn’t use components of the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$.)

Note that vectors $\mathbf{u}, \mathbf{v}\in \R^3$ is said to be perpendicular if
\[\mathbf{u}\cdot \mathbf{v}=\mathbf{u}^{\trans}\mathbf{v}=0.\]

Thus, the result of the problem says that for a fixed vector $\mathbf{u}\in \R^3$, the set of vectors $\mathbf{v}$ that are perpendicular to $\mathbf{u}$ is a subspace in $\R^3$.
(Note that we appy the problem to $\mathbf{b}=\mathbf{u}^{\trans}$.)

The post The Set of Vectors Perpendicular to a Given Vector is a Subspace first appeared on Problems in Mathematics.

(a) Prove that the column vectors of every $3\times 5$ matrix $A$ are linearly dependent.

(b) Prove that the row vectors of every $5\times 3$ matrix $B$ are linearly dependent.

Proof.

(a) Prove that the column vectors of every $3\times 5$ matrix $A$ are linearly dependent.

Note that the column vectors of the matrix $A$ are linearly dependent if the matrix equation
\[A\mathbf{x}=\mathbf{0}\] has a nonzero solution $\mathbf{x}\in \R^5$.

The equation is equivalent to a $3\times 5$ homogeneous system.
As there are more variables than equations, the homogeneous system has infinitely many solutions.

In particular, the equation has a nonzero solution $\mathbf{x}$.
Hence the column vectors are linearly dependent.

(b) Prove that the row vectors of every $5\times 3$ matrix $B$ are linearly dependent.

Observe that the row vectors of the matrix $B$ are the column vectors of the transpose $B^{\trans}$. Note that the size of $B^{\trans}$ is $3\times 5$.

In part (a), we showed that the column vectors of any $3\times 5$ matrix are linearly dependent.
It follows that the column vectors of $B^{\trans}$ are linearly dependent.
Hence the row vectors of $B$ are linearly dependent.

The post The Column Vectors of Every \times 5$ Matrix Are Linearly Dependent first appeared on Problems in Mathematics.

Let $A$ be an $m\times n$ matrix. Prove that the rank of $A$ is the same as the rank of the transpose matrix $A^{\trans}$.

Hint.

Recall that the rank of a matrix $A$ is the dimension of the range of $A$.
The range of $A$ is spanned by the column vectors of the matrix $A$.

Proof.

We write $A=(a_{ij})$ and let
\[\mathbf{A}_i=\begin{bmatrix}
a_{1i} \\
a_{2i} \\
\cdots \\
a_{mi}
\end{bmatrix}\] be the $i$-th column vector of $A$ for $i=1,2,\dots, n$.
Also let
\[\mathbf{B}_i=\begin{bmatrix}
a_{i1} \\
a_{i2} \\
\cdots \\
a_{in}
\end{bmatrix}\] be the $i$-th column vector of $A^{\trans}$ for $i=1, 2, \dots, m$, that is $B_i$ is the transpose of the $i$-th row vector of $A$.

Suppose that $\rk(A)=\dim(\calR(A^{\trans}))=k$ and let $\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$ be a basis for the range $\calR(A^{\trans})$.
We write
\[\mathbf{v}_{i}=\begin{bmatrix}
v_{i1} \\
v_{i2} \\
\vdots \\
v_{in}
\end{bmatrix}\] for $i=1,2, \dots, k$.
Then each column vector $\mathbf{B}_i$ of $A^{\trans}$ is a linear combination of $\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$. Thus we have
\begin{align*}
\mathbf{B}_1 &=c_{11}\mathbf{v}_1+\cdots+c_{1k}\mathbf{v}_k\\
\mathbf{B}_2 &=c_{21}\mathbf{v}_1+\cdots+c_{2k}\mathbf{v}_k\\
\vdots\\
\mathbf{B}_m &=c_{m1}\mathbf{v}_1+\cdots+c_{mk}\mathbf{v}_k.
\end{align*}

More explicitly we have,
\begin{align*}
\begin{bmatrix}
a_{11} \\
a_{12} \\
\cdots \\
a_{1n}
\end{bmatrix} &=c_{11}\begin{bmatrix}
v_{11} \\
v_{12} \\
\vdots \\
v_{1n}
\end{bmatrix}+\cdots+c_{1k}\begin{bmatrix}
v_{k1} \\
v_{k2} \\
\vdots \\
v_{kn}
\end{bmatrix}\\
\begin{bmatrix}
a_{21} \\
a_{22} \\
\cdots \\
a_{2n}
\end{bmatrix} &=c_{21}\begin{bmatrix}
v_{11} \\
v_{12} \\
\vdots \\
v_{1n}
\end{bmatrix}+\cdots+c_{2k}\begin{bmatrix}
v_{k1} \\
v_{k2} \\
\vdots \\
v_{kn}
\end{bmatrix}\\
\vdots\\
\begin{bmatrix}
a_{m1} \\
a_{m2} \\
\cdots \\
a_{mn}
\end{bmatrix} &=c_{m1}\begin{bmatrix}
v_{11} \\
v_{12} \\
\vdots \\
v_{1n}
\end{bmatrix}+\cdots+c_{mk}\begin{bmatrix}
v_{k1} \\
v_{k2} \\
\vdots \\
v_{kn}
\end{bmatrix}\\
\end{align*}

Now, we look at the $i$-th entries for the above vectors and we have
\begin{align*}
a_{1i}&=c_{11}v_{1i}+\cdots+c_{1k}v_{ki}\\
a_{2i}&=c_{21}v_{1i}+\cdots+c_{2k}v_{ki}\\
\vdots \\
a_{mi}&=c_{m1}v_{1i}+\cdots+c_{mk}v_{ki}\\.
\end{align*}
We rewrite these as a vector equality and obtain
\begin{align*}
\mathbf{A}_i&=\begin{bmatrix}
a_{1i} \\
a_{2i} \\
\cdots \\
a_{mi}
\end{bmatrix}=v_{1i}\begin{bmatrix}
c_{11} \\
c_{21} \\
\vdots \\
c_{m1}
\end{bmatrix}+\cdots +v_{ki}\begin{bmatrix}
c_{1k} \\
c_{2k} \\
\vdots \\
c_{mk}
\end{bmatrix}\\
&=v_{1i}\mathbf{c}_{1}+\cdots+v_{ki}\mathbf{c}_k,
\end{align*}
where we put
\[c_j=\begin{bmatrix}
c_{1j} \\
c_{2j} \\
\vdots \\
c_{mj}
\end{bmatrix}\] for $j=1,2,\dots,k$.

This shows that any column vector $\mathbf{A}_i$ of $A$ is a linear combination of vectors $\mathbf{c}_1, \dots, \mathbf{c}_k$. Therefore we have
\[\calR(A)=\Span\{\mathbf{A}_1, \dots, \mathbf{A}_n\} \subset \Span\{\mathbf{c}_1, \dots, \mathbf{c}_k\}. \] Since the dimension of a subspace is smaller than or equal to the dimension of a vector space containing it, we have
\[\rk(A)=\dim(\calR(A)) \leq \dim(\Span\{\mathbf{c}_1, \dots, \mathbf{c}_k\}) \leq k. \] Hence we obtain
\[\rk(A) \leq \rk(A^{\trans}).\]

To achieve the opposite inequality, we repeat this argument using $A^{\trans}$, and we obtain
\[\rk(A^{\trans}) \leq \rk((A^{\trans})^{\trans})=\rk(A)\] since we have $(A^{\trans})^{\trans}=A$.

Therefore, we proved the required equality
\[\rk(A)=\rk(A^{\trans}).\] Click here if solved 31

The post Column Rank = Row Rank. (The Rank of a Matrix is the Same as the Rank of its Transpose) first appeared on Problems in Mathematics.