Let $V$ be a subspace of $\R^n$.
Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$ is a basis of the subspace $V$.

Prove that every basis of $V$ consists of $k$ vectors in $V$.

Hint.

You may use the following fact:

For a proof of this fact, see the post ↴
If there are More Vectors Than a Spanning Set, then Vectors are Linearly Dependent

Proof.

Let $B’=\{\mathbf{w}_1, \mathbf{w}_2, \dots, \mathbf{w}_l\}$ be an arbitrary basis of the subspace $V$.
Our goal is to show that $l=k$.

As $B$ is a basis, it is a spanning set for $V$ consisting of $k$ vectors.
By the fact stated above, a set of $k+1$ or more vectors of $V$ must be linearly dependent.
Since $B’$ is a basis, it is linearly independent.
It follows that $l\leq k$.

We now change the roles of $B$ and $B’$.
As $B’$ is a basis, it is a spanning set for $V$ consisting of $l$ vectors.
So it follows from Fact that a set of $l+1$ or more vectors must be linearly dependent.
Since $B$ is a basis, it is linearly independent.
Hence $k \leq l$.

Therefore we have $l\leq k$ and $k \leq l$, and it yields that $l=k$, as required.

Let $B=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ be a set of three-dimensional vectors in $\R^3$.

(a) Prove that if the set $B$ is linearly independent, then $B$ is a basis of the vector space $\R^3$.

(b) Prove that if the set $B$ spans $\R^3$, then $B$ is a basis of $\R^3$.

Definition (A Basis of a Subspace).

A subset $B$ of a vector space $V$ is called a basis if $B$ is linearly independent spanning set.

Proof.

(a) Prove that if the set $B$ is linearly independent, then $B$ is a basis of the vector space $\R^3$.

To show that $B$ is a basis, we need only prove that $B$ is a spanning set of $\R^3$ as we know that $B$ is linearly independent.
Let $\mathbf{b}\in \R^3$ be an arbitrary vector.
We prove that there exist $x_1, x_2, x_3$ such that
\[x_1\mathbf{v}_1+x_2\mathbf{v}_2+x_3\mathbf{v}_3=\mathbf{b}.\] This is equivalent to having a solution $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ to the matrix equation
\[A\mathbf{x}=\mathbf{b}, \tag{*}\] where
\[A=[\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3]\] is the $3\times 3$ matrix whose column vectors are $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$.

Since the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent, the matrix $A$ is nonsingular.
It follows that the equation (*) has the unique solution $\mathbf{x}=A^{-1}\mathbf{b}$.
Hence $\mathbf{b}$ is a linear combination of the vectors in $B$.
This means that $B$ is a spanning set of $\R^3$, hence $B$ is a basis.

(b) Prove that if the set $B$ spans $\R^3$, then $B$ is a basis of $\R^3$.

As we know that $B$ spans $\R^3$, it suffices to show that $B$ is linearly independent.
Note that the assumption that $B$ is a spanning set of $\R^3$ means that any vector $\mathbf{b}$ in $\R^3$ is a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$: there exist $c_1, c_2, c_3$ such that
\[c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{b}.\] Equivalently, for any $\mathbf{b}\in \R^3$, the equation $A\mathbf{x}=\mathbf{b}$ has a solution.

Let $A=[\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3]$ as before.
It follows that the equation
\[A\mathbf{x}=\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}\] has a solution $\mathbf{x}=\mathbf{u}_1$.
Similarly the equations
\[A\mathbf{x}=\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}, \quad A\mathbf{x}=\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\] have solutions $\mathbf{u}_2, \mathbf{u}_3$, respectively.

Define the $3\times 3$ matrix $A’$ by
\[A’=[\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3].\]

Then it follows that
\begin{align*}
AA’&=A[\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3]\\
&=[A\mathbf{u}_1, A\mathbf{u}_2, A\mathbf{u}_3]\\
&=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}

As the identity matrix is nonsingular, the product $AA’$ is nonsingular.
Thus, the matrix $A$ is nonsingular as well.
This implies that the column vectors of $A$ are linearly independent.
Hence the set $B$ is linearly independent and we conclude that $B$ is a basis of $\R^3$.

Related Question.

Use the result of the problem, try the next problem about a basis for $\R^3$.

Let $V$ be a vector space over the field of real numbers $\R$.

Prove that if the dimension of $V$ is $n$, then $V$ is isomorphic to $\R^n$.

Proof.

Since $V$ is an $n$-dimensional vector space, it has a basis
\[B=\{\mathbf{v}_1, \dots, \mathbf{v}_n\},\] where each $\mathbf{v}_i$ is a vector in $V$.

Define a map $T: V\to \R^n$ by sending each vector $\mathbf{v}\in V$ to its coordinate vector $[\mathbf{v}]_B$ with respect to the basis $B$.
More explicitly, if
\[\mathbf{v}=c_1\mathbf{v}_1+\cdots+c_n \mathbf{v}_n \text{ with } c_1, \dots, c_n \in \R,\] then the coordinate vector with respect to $B$ is
\[[\mathbf{v}]_B=\begin{bmatrix}
c_1 \\
c_2 \\
\vdots \\
c_n
\end{bmatrix} \in \R^n.\] Then the map $T: V \to \R^n$ is defined by
\[T(\mathbf{v})=\begin{bmatrix}
c_1 \\
c_2 \\
\vdots \\
c_n
\end{bmatrix}.\]

It follows from the properties of the coordinate vectors that the map $T$ is a linear transformation.
We show that $T$ is bijective, hence an isomorphism.

$T$ is injective.

To show that $T$ is injective, it suffices to show that the null space of $T$ is trivial: $\calN(T)=\{\mathbf{0}\}$.
(See the post “A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero” for a proof of this fact.)

If $\mathbf{v}\in \calN(T)$, then we have
\[\mathbf{0}=T(\mathbf{v})=[\mathbf{v}]_B.\] So the coordinate vector of $\mathbf{v}$ is zero, hence we have
\[\mathbf{v}=0\mathbf{v}_1+\cdots +0\mathbf{v}_n=\mathbf{0}.\] Thus, $\calN(T)=\{\mathbf{0}\}$, and $T$ is injective.

$T$ is surjective.

To show that $T$ is surjective, let
\[\mathbf{a}=\begin{bmatrix}
a_1 \\
a_2 \\
\vdots \\
a_n
\end{bmatrix}\] be an arbitrary vector in $\R^n$.
Then consider the vector
\[\mathbf{v}:=a_1\mathbf{v}_1+\cdots+ a_n \mathbf{v}_n\] in $V$.
Then it follows from the definition of the linear transformation $T$ that
\[T(\mathbf{v})=[\mathbf{v}]_B=\begin{bmatrix}
a_1 \\
a_2 \\
\vdots \\
a_n
\end{bmatrix}=\mathbf{a}.\] Therefore $T$ is surjective.

In summary, $T: V \to \R^n$ is a bijective linear transformation, and hence $T$ is an isomorphism.
Thus, we conclude that $V$ and $\R^n$ are isomorphic.

Generalization

We may use more general field $\F$ instead of the field of real numbers $\R$.
Then the general statement is as follows.

The proof is identical as above except that we replace $\R$ by $\F$ everywhere.

Consider the $2\times 2$ real matrix
\[A=\begin{bmatrix}
1 & 1\\
1& 3
\end{bmatrix}.\]

(a) Prove that the matrix $A$ is positive definite.

(b) Since $A$ is positive definite by part (a), the formula
\[\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans} A \mathbf{y}\] for $\mathbf{x}, \mathbf{y} \in \R^2$ defines an inner product on $\R^n$.
Consider $\R^2$ as an inner product space with this inner product.

Prove that the unit vectors
\[\mathbf{e}_1=\begin{bmatrix}
1 \\
0
\end{bmatrix} \text{ and } \mathbf{e}_2=\begin{bmatrix}
0 \\
1
\end{bmatrix}\] are not orthogonal in the inner product space $\R^2$.

(c) Find an orthogonal basis $\{\mathbf{v}_1, \mathbf{v}_2\}$ of $\R^2$ from the basis $\{\mathbf{e}_1, \mathbf{e}_2\}$ using the Gram-Schmidt orthogonalization process.

Proof.

(a) Prove that the matrix $A$ is positive definite.

We prove that for every nonzero vector $\mathbf{x}=\begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2$, we have $\mathbf{x}^{\trans} A \mathbf{x} > 0$.
We have
\begin{align*}
\mathbf{x}^{\trans} A \mathbf{x}&=\begin{bmatrix}
x & y
\end{bmatrix} \begin{bmatrix}
1 & 1\\
1& 3
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
x & y
\end{bmatrix}\begin{bmatrix}
x+y \\
x+3y
\end{bmatrix}\\[6pt] &=x(x+y)+y(x+3y)=x^2+2xy+3y^2\\
&=x^2+2xy+y^2+2y^2=(x+y)^2+2y^2.
\end{align*}

Since $\mathbf{x}\neq \mathbf{0}$, at least one of $x, y$ is nonzero.
Thus the last expression is always positive.
Hence $A$ is a positive definite matrix.

(b) Prove that $\mathbf{e}_1, \mathbf{e}_2$ are not orthogonal in the inner product space $\R^2$.

Note that by post “A Symmetric Positive Definite Matrix and An Inner Product on a Vector Space“, the formula $\langle \mathbf{x}, \mathbf{y}\rangle$ defines an inner product on $\R^2$.

Two vectors $\mathbf{x}$ and $\mathbf{y}$ is said to be orthogonal if $\langle \mathbf{x}, \mathbf{y}\rangle=0$.

The vectors $\mathbf{e}_1, \mathbf{e}_2$ are not orthogonal with this inner product since
\begin{align*}
\langle \mathbf{e}_1, \mathbf{e}_2\rangle=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
1 & 1\\
1& 3
\end{bmatrix}\begin{bmatrix}
0 \\
1
\end{bmatrix}=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
1 \\
3
\end{bmatrix}=1\neq 0.
\end{align*}

(c) Find an orthogonal basis using the Gram-Schmidt orthogonalization process.

By the Gram-Schmidt orthogonalization process, we have
\begin{align*}
\mathbf{v}_1&=\mathbf{e}_1\\
\mathbf{v}_2&=\mathbf{e}_2-\frac{\langle \mathbf{v}_1, \mathbf{e}_2 \rangle}{\langle \mathbf{v}_1, \mathbf{v}_1 \rangle}\mathbf{v}_1
=\mathbf{e}_2-\frac{\langle \mathbf{e}_1, \mathbf{e}_2 \rangle}{\langle \mathbf{e}_1, \mathbf{e}_1 \rangle}\mathbf{e}_1.
\end{align*}

We compute
\begin{align*}
\langle \mathbf{e}_1, \mathbf{e}_1 \rangle=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
1 & 1\\
1& 3
\end{bmatrix}\begin{bmatrix}
1 \\
0
\end{bmatrix}=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
1 \\
1
\end{bmatrix}=1.
\end{align*}
We also have $\langle \mathbf{e}_1, \mathbf{e}_2\rangle=1$ from part (b).
Thus, we have
\begin{align*}
\mathbf{v}_2=\mathbf{e}_2-\mathbf{e}_1=\begin{bmatrix}
-1 \\
1
\end{bmatrix}.
\end{align*}
Thus, the Gram-Schmidt orthogonalization process yields the orthogonal basis
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
0
\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}
-1 \\
1
\end{bmatrix}.\]

Double Check

Let us verify that $\mathbf{v}_1, \mathbf{v}_2$ are orthogonal by computing their inner product directly as follows.
We have
\begin{align*}
\langle \mathbf{v}_1, \mathbf{v}_2\rangle=\mathbf{v}_1^{\trans} A\mathbf{v}_2=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
1 & 1\\
1& 3
\end{bmatrix}\begin{bmatrix}
-1 \\
1
\end{bmatrix}=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
0 \\
2
\end{bmatrix}=0.
\end{align*}

Determine all linear transformations of the $2$-dimensional $x$-$y$ plane $\R^2$ that take the line $y=x$ to the line $y=-x$.

Solution.

Let $T:\R^2 \to \R^2$ be a linear transformation that maps the line $y=x$ to the line $y=-x$.
Note that the linear transformation $T$ is completely determined if the values of $T$ on basis vectors of the vector space $\R^2$ are known.

Let
\[B=\left\{\, \begin{bmatrix}
1 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
1
\end{bmatrix} \,\right\}\] be a basis of $\R^2$.

The reason of this choice is as follows.
Since we know that $T$ takes the line $y=x$ to the line $y=-x$, the vector $\begin{bmatrix}
1 \\
1
\end{bmatrix}$ is mapped into some point on the line $y=-x$.
This is how we chose the vector $\begin{bmatrix}
1 \\
1
\end{bmatrix}$. The other basis vector could be any vector that is not a multiple of $\begin{bmatrix}
1 \\
1
\end{bmatrix}$, and we just chose the simple vector $\begin{bmatrix}
1 \\
0
\end{bmatrix}$.

Let
\[T\left(\, \begin{bmatrix}
1 \\
0
\end{bmatrix} \,\right)=\begin{bmatrix}
a \\
b
\end{bmatrix}\] for some $a,b \in \R$.

Since we know that the vector $T\left(\, \begin{bmatrix}
1 \\
1
\end{bmatrix} \,\right)$ is on the line $y=-x$, let
\[T\left(\, \begin{bmatrix}
1 \\
1
\end{bmatrix} \,\right)=\begin{bmatrix}
c \\
-c
\end{bmatrix}\] for some $c\in \R$.

We now find a formula for this linear transformation $T$.
Let $\begin{bmatrix}
x \\
y
\end{bmatrix}$ be an arbitrary vector in the plane $\R^2$.

We express this vector as a linear combination of basis vectors:
\[\begin{bmatrix}
x \\
y
\end{bmatrix}=(x-y)\begin{bmatrix}
1 \\
0
\end{bmatrix}+y\begin{bmatrix}
1 \\
1
\end{bmatrix}.\]

Then we have
\begin{align*}
&T\left(\, \begin{bmatrix}
x \\
y
\end{bmatrix} \,\right)\\
&=T\left(\, (x-y)\begin{bmatrix}
1 \\
0
\end{bmatrix}+y\begin{bmatrix}
1 \\
1
\end{bmatrix} \,\right)\\
&=(x-y)T\left(\, \begin{bmatrix}
1 \\
0
\end{bmatrix}\,\right)+yT\left(\, \begin{bmatrix}
1 \\
1
\end{bmatrix} \,\right) && \text{since $T$ is a linear transformation}\\
&=(x-y)\begin{bmatrix}
a \\
b
\end{bmatrix}+y\begin{bmatrix}
c \\
-c
\end{bmatrix}\\
&=\begin{bmatrix}
ax+(c-a)y \\
bx-(c+b)y
\end{bmatrix}.
\end{align*}

We conclude that any linear transformation $T:\R^2\to \R^2$ that takes the line $y=x$ to the line $y=-x$ is of the form

for some $a, b, c\in \R$.

Remark.

Remark that if $c=0$, then all the points on the line $y=x$ are mapped into the origin, which is on the line $y=-x$.
If we want to avoid this degenerate case, we need to assume that $c\neq 0$.

Let $U$ and $V$ be finite dimensional subspaces in a vector space over a scalar field $K$.
Then prove that
\[\dim(U+V) \leq \dim(U)+\dim(V).\]

Definition (The sum of subspaces).

Recall that the sum of subspaces $U$ and $V$ is
\[U+V=\{\mathbf{x}+\mathbf{y} \mid \mathbf{x}\in U, \mathbf{y}\in V\}.\] The sum $U+V$ is a subspace.
(See the post “The sum of subspaces is a subspace of a vector space” for a proof.)

Proof.

Let $n=\dim(U)$ and $m=\dim(V)$.
Let
\[B_1=\{\mathbf{u}_1, \dots, \mathbf{u}_n\}\] be a basis of the vector space $U$ and let
\[B_2=\{\mathbf{v}_1, \dots, \mathbf{v}_m\}\] be a basis of the vector space $V$.

An arbitrary element of the vector space $U+W$ is of the form $\mathbf{x}+\mathbf{y}$, where $\mathbf{x}\in U$ and $\mathbf{y} \in V$.

Since $B_1$ is a basis of $U$, we can write
\[\mathbf{x}=r_1\mathbf{u}_1+\cdots +r_n \mathbf{u}_n\] for some scalars $r_1, \dots, r_n\in K$.
Also, since $B_2$ is a basis of $V$, we can write
\[\mathbf{y}=s_1\mathbf{v}_1+\cdots +s_m \mathbf{v}_m\] for some scalars $s_1, \dots, s_m\in K$.

Thus, we have
\begin{align*}
\mathbf{x}+\mathbf{y}&=r_1\mathbf{u}_1+\cdots +r_n \mathbf{u}_n+s_1\mathbf{v}_1+\cdots +s_m \mathbf{v}_m,
\end{align*}
and hence $\mathbf{x}+\mathbf{y}$ is in the span $S:=\Span(\mathbf{u}_1, \dots, \mathbf{u}_n, \mathbf{v}_1, \dots, \mathbf{v}_m)$.

Thus we have $U+W \subset S$ and it yields that \begin{align*}
\dim(U+W) \leq \dim(S)\leq n+m=\dim(U)+\dim(V).
\end{align*}
This completes the proof.

Related Question.

See the post ↴
The rank of the sum of two matrices
for a proof of this problem.

Let $n>1$ be a positive integer. Let $V=M_{n\times n}(\C)$ be the vector space over the complex numbers $\C$ consisting of all complex $n\times n$ matrices. The dimension of $V$ is $n^2$.
Let $A \in V$ and consider the set
\[S_A=\{I=A^0, A, A^2, \dots, A^{n^2-1}\}\] of $n^2$ elements.
Prove that the set $S_A$ cannot be a basis of the vector space $V$ for any $A\in V$.

Proof.

We prove that the set $S_A$ is linearly dependent, hence it cannot be a basis of $V$.
Since $A$ is an $n\times n$ matrix, its characteristic polynomial $p(t)=\det(tI-A)$ is a degree $n$ polynomial.

(Your preferred definition of the characteristic polynomial might be $\det(A-tI)$. It is straight forward to modify the following proof with this definition.)

Let us write it as
\[p(t)=t^n+a_{n-1}t^{n-1}+\cdots+a_1x+a_0.\] Then the Cayley-Hamilton theorem states that
\[p(A)=A^n+a_{n-1}A^{n-1}+\cdots+a_1A+a_0I=O\] is the zero matrix.

Since the coefficient of $A^n$ is $1$, this gives a non-trivial linear combination of $I, A, \dots, A^n$. Therefore the set
\[T:=\{I, A, \dots, A^n\}\] is linearly dependent.

As $T$ is a subset of $S_A$, the set $S_A$ is also linearly dependent.
Therefore, $S_A$ is not a basis of $V$. This completes the proof.

Let $V$ be a vector space over $\R$ and let $B$ be a basis of $V$.
Let $S=\{v_1, v_2, v_3\}$ be a set of vectors in $V$. If the coordinate vectors of these vectors with respect to the basis $B$ is given as follows, then find the dimension of $V$ and the dimension of the span of $S$.
\[[v_1]_B=\begin{bmatrix}
1 \\
0 \\
0 \\
0
\end{bmatrix}, [v_2]_B=\begin{bmatrix}
0 \\
1 \\
0 \\
0
\end{bmatrix}, [v_3]_B=\begin{bmatrix}
1 \\
1 \\
0 \\
0
\end{bmatrix}.\]

Solution.

Coordinate vectors

Let us first recall the definition of coordinate vectors.
Suppose that $V$ is a vector space over $\R$ and let $B=\{v_1, v_2, \dots, v_n\}$ be a basis of $V$ (hence the dimension of $V$ is $n$).

Then any element of $v$ of $V$ can be written as
\[v=c_1v_1+c_2v_2+\cdots +c_n v_n,\] where $c_1, c_2, \dots, c_n$ are scalars in $\R$.

This expression is unique and the coordinate vector of $v$ with respect to the basis $B$ is defined as
\[[v]_B=\begin{bmatrix}
c_1 \\
c_2 \\
\vdots \\
c_n
\end{bmatrix}.\] Thus the coordinate vector $[v]_B$ is just a usual $n$-dimensional vector.

Main part of the solution

Note that in the current problem, the coordinate vectors are $4$-dimensional vectors.
This implies that the basis $B$ consists of four vectors. Hence the dimension of $V$ is $4$.

By the correspondence of the coordinate vectors, the dimension of $\Span(S)$ is the same as the dimension of $\Span(T)$, where
\begin{align*}
T&=\{[v_1]_B, [v_2]_B, [v_2]_B\}\\
&=\left\{\, \begin{bmatrix}
1 \\
0 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
1 \\
0 \\
0
\end{bmatrix} \,\right\}.
\end{align*}

To find the dimension of $\Span(T)$, we need to find a basis of $\Span(T)$.
One way to do this is to note that the third vector is the sum of the first two vectors. Also, it’s clear that the first two vectors are linearly independent.

Thus, the set
\begin{align*}
\left\{\, \begin{bmatrix}
1 \\
0 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
0 \\
0
\end{bmatrix} \,\right\}
\end{align*}
is a basis of $\Span(T)$, hence the dimension of $\Span(T)$ is $2$.
We conclude that the dimension of $\Span(S)$ is $2$ as well.
(We can also conclude that the set $\{v_1, v_2\}$ is a basis of $\Span(S)$.)

Another way to find a basis of $\Span(T)$

Here is another way to find a basis of $\Span(T)$. We can use the leading 1 method.
We form the matrix whose column vectors are the vectors in $T$:
\begin{align*}
\begin{bmatrix}
1 & 0 & 1 \\
0 &1 &1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
\end{align*}

Note that this matrix is already in reduced row echelon form.
The first two columns contain the leading 1’s. Thus the first two columns form a basis of $\Span(T)$ (this is the leading 1 method).

Remark: In general we apply elementary row operations to reduce the matrix into a matrix in reduced row echelon form.

Let $V$ be a subset of the vector space $\R^n$ consisting only of the zero vector of $\R^n$. Namely $V=\{\mathbf{0}\}$.
Then prove that $V$ is a subspace of $\R^n$.

Proof.

To prove that $V=\{\mathbf{0}\}$ is a subspace of $\R^n$, we check the following subspace criteria.

Condition (a) is clear since $V$ consists of the zero vector $\mathbf{0}$.

To check condition (b), note that the only element in $V=\{\mathbf{0}\}$ is $\mathbf{0}$. Thus if $\mathbf{x}, \mathbf{y} \in V$, then both $\mathbf{x}, \mathbf{y}$ are $\mathbf{0}$. Hence
\[\mathbf{x}+\mathbf{y} =\mathbf{0}+\mathbf{0}=\mathbf{0}\in V\] and condition (b) is met.

To confirm condition (c), let $\mathbf{x}\in V$ and $c\in \R$. Then $\mathbf{x}=\mathbf{0}$.
We have
\[c\mathbf{x}=c\mathbf{0}=\mathbf{0}\in V\] and condition (c) is satisfied.

Hence we have checked all the subspace criteria, and hence the subset $V=\{\mathbf{0}\}$ consisting only of the zero vector is a subspace of $\R^n$.

What’s the dimension of the zero vector space?

What’s the dimension of the subspace $V=\{\mathbf{0}\}$?

The dimension of a subspace is the number of vectors in a basis. So let us first find a basis of $V$.

Note that a basis of $V$ consists of vectors in $V$ that are linearly independent spanning set. Since $0$ is the only vector in $V$, the set $S=\{\mathbf{0}\}$ is the only possible set for a basis.

However, $S$ is not a linearly independent set since, for example, we have a nontrivial linear combination $1\cdot \mathbf{0}=\mathbf{0}$.

Therefore, the subspace $V=\{\mathbf{0}\}$ does not have a basis.
Hence the dimension of $V$ is zero.

Let $V$ be the vector space of all $3\times 3$ real matrices.
Let $A$ be the matrix given below and we define
\[W=\{M\in V \mid AM=MA\}.\] That is, $W$ consists of matrices that commute with $A$.
Then $W$ is a subspace of $V$.

Determine which matrices are in the subspace $W$ and find the dimension of $W$.

(a) \[A=\begin{bmatrix}
a & 0 & 0 \\
0 &b &0 \\
0 & 0 & c
\end{bmatrix},\] where $a, b, c$ are distinct real numbers.

(b) \[A=\begin{bmatrix}
a & 0 & 0 \\
0 &a &0 \\
0 & 0 & b
\end{bmatrix},\] where $a, b$ are distinct real numbers.

Solution.

(a) Diagonal matrix with distinct diagonal entries

Let us first determine when a matrix $M$ commutes with $A$.
Let
\[M=\begin{bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end{bmatrix}\] and suppose that $AM=MA$:
\[\begin{bmatrix}
a & 0 & 0 \\
0 & b &0 \\
0 & 0 & c
\end{bmatrix}
\begin{bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end{bmatrix}
=
\begin{bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end{bmatrix}
\begin{bmatrix}
a & 0 & 0 \\
0 &b &0 \\
0 & 0 & c
\end{bmatrix}.\] Computing matrix products, we obtain
\[\begin{bmatrix}
aa_{1 1} & aa_{1 2} & aa_{1 3} \\
ba_{2 1} & ba_{2 2} & ba_{2 3} \\
ca_{3 1} & ca_{3 2} &c a_{3 3}
\end{bmatrix}
=
\begin{bmatrix}
a_{1 1}a & a_{1 2}b & a_{1 3}c \\
a_{2 1}a & a_{2 2}b & a_{2 3}c\\
a_{3 1}a & a_{3 2}b & a_{3 3}c
\end{bmatrix}. \tag{*}\] Compare the $(1,2)$ entries and we have $aa_{1 2}=ba_{1 2}$.
Since $a\neq b$, we must have $a_{1 2}=0$.

Similarly, comparing the off-diagonal entries and noting $a, b, c$ are distinct, we find that off diagonal entries $a_{i j} , i\neq j$ must be $0$.

Thus, $M$ commutes with $A$ if and only if
\[M=\begin{bmatrix}
a_{1 1} & 0 & 0 \\
0 & a_{2 2} & 0 \\
0 & 0 & a_{3 3}
\end{bmatrix}.\]

Therefore, the subspace $W$ consists of all $3\times 3$ diagonal matrices:
\[W=\{W\in V\mid W \text{ is diagonal}\}.\] Then it is easy to see that the set $\{E_{1 1}, E_{2 2}, E_{3 3}\}$ is a basis for $W$, where $E_{i j}$ is the $3\times 3$ matrix whose $(i,j)$-entry is $1$ and the other entries are zero. Thus the dimension of $W$ is $3$.

(b) Diagonal matrix two diagonal entries are the same

Now consider the case
\[A=\begin{bmatrix}
a & 0 & 0 \\
0 &a &0 \\
0 & 0 & b
\end{bmatrix}.\] Let
\[M=\begin{bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end{bmatrix}\] and compute $AM=MA$ as in part (a) (or you just need to replace $b, c$ in (*) by $a, b$, respectively) and obtain
\[\begin{bmatrix}
aa_{1 1} & aa_{1 2} & aa_{1 3} \\
aa_{2 1} & aa_{2 2} & aa_{2 3} \\
ba_{3 1} & ba_{3 2} & ba_{3 3}
\end{bmatrix}
=
\begin{bmatrix}
a_{1 1}a & a_{1 2}a & a_{1 3}b \\
a_{2 1}a & a_{2 2}a & a_{2 3}b\\
a_{3 1}a & a_{3 2}a & a_{3 3}b
\end{bmatrix}. \] Comparing entries and noting $a\neq b$, we have
\[a_{1 3}=0, a_{2 3}=0, a_{3 1}=0, a_{3 2}=0.\]

Thus, $M$ commutes with $A$ is and only if
\[M=\begin{bmatrix}
a_{1 1} & a_{1 2} & 0 \\
a_{2 1} & a_{2 2} & 0 \\
0 & 0 & a_{3 3}
\end{bmatrix},\] and hence the subspace $W$ consists of such matrices.
From this, we see that the set
\[\{E_{1 1}, E_{1 2}, E_{2 1}, E_{2 2}, E_{3 3}\}\] is a basis for $W$, and we conclude that the dimension of $W$ is $5$.