Let $V$ be the vector space of $k \times k$ matrices. Then for fixed matrices $R, S \in V$, define the subset $W = \{ R A S \mid A \in V \}$.

Prove that $W$ is a vector subspace of $V$.

Proof.

We verify the subspace criteria: the zero vector of $V$ is in $W$, and $W$ is closed under addition and scalar multiplication.

First, let $\mathbf{0} \in V$ be the $k\times k$ zero matrix. Then $ R \mathbf{0} S = \mathbf{0}$, and so $\mathbf{0} \in W$.

Now suppose $X, Y \in W$. Then there are elements $A, B \in V$ such that $RAS = X$ and $RBS = Y$. Then
\[X + Y = RAS + RBS = R (A+B) S\] and so $X+Y \in W$.

Now for a scalar $c \in \mathbb{R}$ and matrix $X = RAS \in W$, we have
\[cX = c RAS = R (cA )S,\] and so $cX \in W$ as well.

The post For Fixed Matrices $R, S$, the Matrices $RAS$ form a Subspace first appeared on Problems in Mathematics.

For what real values of $a$ is the set
\[W_a = \{ f \in C(\mathbb{R}) \mid f(0) = a \}\] a subspace of the vector space $C(\mathbb{R})$ of all real-valued functions?

Solution.

The zero element of $C(\mathbb{R})$ is the function $\mathbf{0}$ defined by $\mathbf{0}(x) = 0$. This shows that $\mathbf{0} \in W_a$ if and only if $a=0$.

We have shown that if $a \neq 0$, then $W_a$ is not a subspace as every subspace contains the zero vector. Now we consider the case $a=0$ and prove that $W_0$ is a subspace.

We verify the subspace criteria: the zero vector of $C(\R)$ is in $W_0$, and $W_0$ is closed under addition and scalar multiplication.

As mentioned before, $\mathbf{0} \in W_0$.

Now suppose $f, g \in W_0$. Then $f(0) = g(0) = 0$, and so
\[(f+g)(0) = f(0) + g(0) = 0.\] Thus $f+g \in W_0$. Finally, if $c \in \mathbb{R}$ is a scalar and $f \in W_0$, then
\[(cf)(0) = c f(0) = c \cdot 0 = 0.\] Thus $cf \in W_0$, and $W_0$ is a vector subspace.

The post Determine the Values of $a$ so that $W_a$ is a Subspace first appeared on Problems in Mathematics.

Let $V$ be the vector space of $n \times n$ matrices, and $M \in V$ a fixed matrix. Define
\[W = \{ A \in V \mid AM = MA \}.\] The set $W$ here is called the centralizer of $M$ in $V$.

Prove that $W$ is a subspace of $V$.

Proof.

First we check that the zero element of $V$ lies in $W$. The zero element of $V$ is the $n \times n$ zero matrix $\mathbf{0}$.

It is clear that $M \mathbf{0} = \mathbf{0} = \mathbf{0} M$, and so $\mathbf{0} \in W$.

Next suppose $A, B \in W$ and $c \in \mathbb{R}$. Then $AM = MA$ and $BM = MB$, and so
\[( A + B ) M = A M + B M = M A + M B = M ( A + B ).\] Thus, $A + B \in W$.

We also have
\[( c A ) M = c ( A M ) = c ( M A ) = M ( c A ),\] and so $c A \in W$.

These three criteria show that $W$ is a subspace of $V$.

The post The Centralizer of a Matrix is a Subspace first appeared on Problems in Mathematics.

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.

Let $V$ be the vector space of $n \times n$ matrices with real coefficients, and define
\[ W = \{ \mathbf{v} \in V \mid \mathbf{v} \mathbf{w} = \mathbf{w} \mathbf{v} \mbox{ for all } \mathbf{w} \in V \}.\] The set $W$ is called the center of $V$.

Prove that $W$ is a subspace of $V$.

Proof.

We must show that $W$ satisfies the three criteria for vector subspaces.
Namely, the zero vector of $V$ is in $W$, and $W$ is closed under addition and scalar multiplication.

First, the zero element in $V$ is the matrix $\mathbf{0}$ whose entries are all $0$. For any other matrix $\mathbf{x} \in V$, we have $\mathbf{0} \mathbf{x} = \mathbf{0} = \mathbf{x} \mathbf{0}$. So we see that $\mathbf{0} \in W$.

Now suppose $\mathbf{v}, \mathbf{w} \in W$ and $c \in \mathbb{R}$. Then for any $\mathbf{x} \in V$, we have
\[(\mathbf{v} + \mathbf{w} ) \mathbf{x} = \mathbf{v} \mathbf{x} + \mathbf{w} \mathbf{x} = \mathbf{x} \mathbf{v} + \mathbf{x} \mathbf{w} = \mathbf{x} ( \mathbf{v} + \mathbf{w} ),\] where the second equality follows because $\mathbf{v}$ and $\mathbf{w}$ lie in $W$. So we see that $\mathbf{v} + \mathbf{w} \in W$ as well, and so $W$ is closed under addition.

Finally we must show that $c \mathbf{v} \in W$ as well. For any other $\mathbf{x} \in V$, we have
\[(c \mathbf{v} ) \mathbf{x} = c ( \mathbf{v} \mathbf{x}) = c ( \mathbf{x} \mathbf{v} ) = \mathbf{x} ( c \mathbf{v} ),\] where the second equality follows from the fact that $\mathbf{v} \in W$ and so $\mathbf{v} \mathbf{x} = \mathbf{x} \mathbf{v}$.

Thus we see that $c \mathbf{v} \in W$, finishing the proof.

The post Prove that the Center of Matrices is a Subspace first appeared on Problems in Mathematics.

Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.
Let
\[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix}
a & b\\
c& -a
\end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\]

(a) Show that $W$ is a subspace of $V$.

(b) Find a basis of $W$.

(c) Find the dimension of $W$.

(The Ohio State University, Linear Algebra Midterm)
 

Solution.

(a) Show that $W$ is a subspace of $V$.

To show that $W$ is a subspace of $V$, we verify the following subspace criteria.

1. The zero vector in $V$ is in $W$.
2. For all $A, B\in W$, the sum $A+B \in W$.
3. For all $A\in W$ and $r\in \R$, the scalar multiplication $rA\in W$.

Note that the zero vector in $V$ is the zero matrix $\begin{bmatrix}
0& 0 \\ 0& 0
\end{bmatrix}$.
The zero matrix corresponds to the matrix in $W$ with $a=b=c=0$, and hence the zero vector of $V$ is in $W$.
Condition 1 is met.

To verify condition 2, let
\[A=\begin{bmatrix}
a & b\\
c& -a
\end{bmatrix} \text{ and } A’=\begin{bmatrix}
a’ & b’\\
c’& -a’
\end{bmatrix}\] be arbitrary elements in $W$, where $a, b, c, d, a’, b’, c’, d’\in \R$.
Then we have
\[A+A’=\begin{bmatrix}
a & b\\
c& -a
\end{bmatrix} +\begin{bmatrix}
a’ & b’\\
c’& -a’
\end{bmatrix}=\begin{bmatrix}
a+a’ & b+b’\\
c+c’& -(a+a)’
\end{bmatrix}\] and this is of the form of elements of $W$. Hence $A+A’\in W$ and condition 2 is met.

Finally, let us check condition 3.
Let $\begin{bmatrix}
a & b\\
c& -a
\end{bmatrix}$ be an arbitrary element in $W$ and let $r\in \R$ be any real number.

Then we have
\begin{align*}
rA=r\cdot \begin{bmatrix}
a & b\\
c& -a
\end{bmatrix}=\begin{bmatrix}
ra & rb\\
rc& -(ra)
\end{bmatrix}
\end{align*}
and this is of the form of the elements of $W$. Thus $rA\in W$ and condition 3 is met.

Therefore, by the subspace criteria, we conclude that $W$ is a subspace of $V$.

(b) Find a basis of $W$.

Any vector $A=\begin{bmatrix}
a & b\\
c& -a
\end{bmatrix}$ in the subspace $W$ can be written as
\[\begin{bmatrix}
a & b\\
c& -a
\end{bmatrix}=a\begin{bmatrix}
1 & 0\\
0& -1
\end{bmatrix}+b\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}+c\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}.\] It follows that the matrices
\[\begin{bmatrix}
1 & 0\\
0& -1
\end{bmatrix}, \begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}, \begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}\] span the subspace $W$.

If the above linear combination is the zero matrix, then $a=b=c=0$.
This implies that these matrices are linearly independent.

Thus,
\[\left\{\, \begin{bmatrix}
1 & 0\\
0& -1
\end{bmatrix}, \begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}, \begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}\,\right\}\] is a basis for $W$.

(c) Find the dimension of $W$.

As the basis we obtained contains three vectors, the dimension of $W$ is $3$.

The post Vector Space of 2 by 2 Traceless Matrices first appeared on Problems in Mathematics.

Let $U$ and $V$ be subspaces of the $n$-dimensional vector space $\R^n$.

Prove that the intersection $U\cap V$ is also a subspace of $\R^n$.

Definition (Intersection).

Recall that the intersection $U\cap V$ is the set of elements that are both elements of $U$ and $V$.
In the set theoretical notation, we have
\[U \cap V=\{x \mid x\in U \text{ and } x\in V\}.\]

Proof.

To prove that the intersection $U\cap V$ is a subspace of $\R^n$, we check the following subspace criteria:

As $U$ and $V$ are subspaces of $\R^n$, the zero vector $\mathbf{0}$ is in both $U$ and $V$.
Hence the zero vector $\mathbf{0}\in \R^n$ lies in the intersection $U \cap V$.
So condition 1 is met.

Suppose that $\mathbf{x}, \mathbf{y} \in U \cap V$.
This implies that $\mathbf{x}$ is a vector in $U$ as well as a vector in $V$.
Similarly, $\mathbf{y}$ is a vector in $U$ as well as a vector in $V$.

Since $U$ is a subspace and $\mathbf{x}$ and $\mathbf{y}$ are both vectors in $U$, their sum $\mathbf{x}+\mathbf{y}$ is in $U$.
Similarly, since $V$ is a subspace and $\mathbf{x}$ and $\mathbf{y}$ are both vectors in $V$, their sum $\mathbf{x}+\mathbf{y}\in V$.

Therefore the sum $\mathbf{x}+\mathbf{y}$ is a vector in both $U$ and $V$.
Hence $\mathbf{x}+\mathbf{y}\in U \cap V$.
Thus condition 2 is met.

To verify condition 3, let $\mathbf{x}\in U \cap V$ and $r\in \R$.
As $\mathbf{x}\in U \cap V$, the vector $\mathbf{x}$ lies in both $U$ and $V$.
Since both $U$ and $V$ are subspaces, the scalar multiplication is closed in $U$ and $V$, respectively.

Thus $r\mathbf{x}\in U$ and $r\mathbf{x}\in V$.
It follows that $r\mathbf{x}\in U\cap V$.

This proves condition 3, and hence the intersection $U\cap V$ is a subspace of $\R^n$.

The post The Intersection of Two Subspaces is also a Subspace first appeared on Problems in Mathematics.

Let $V$ be a vector space over a scalar field $K$.
Let $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ be vectors in $V$ and consider the subset
\[W=\{a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k \mid a_1, a_2, \dots, a_k \in K \text{ and } a_1+a_2+\cdots+a_k=0\}.\] So each element of $W$ is a linear combination of vectors $\mathbf{v}_1, \dots, \mathbf{v}_k$ such that the sum of the coefficients is zero.

Prove that $W$ is a subspace of $V$.

 

We give two proofs.

Proof 1. (Subspace Criteria)

We use the following subspace criteria.
The subset $W$ is a subspace of $V$ if the following three conditions are met.

The zero vector $\mathbf{0}$ of $V$ can be written as
\[\mathbf{0}=0\mathbf{v}_1+0\mathbf{v}_2+\cdots+0\mathbf{v}_k.\] Clearly the sum of the coefficient is zero, hence $\mathbf{0} \in W$.
So condition 1 is met.

To verify condition 2, let
\[\mathbf{v}=a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k\] and
\[\mathbf{v}’=b_1\mathbf{v}_1+b_2\mathbf{v}_2+\cdots+ b_k\mathbf{v}_k\] be arbitrary elements in $W$. Thus
\[a_1+a_2+\cdots+a_k=0 \text{ and } b_1+b_2+\cdots+b_k=0. \tag{*}\] The sum $\mathbf{v}+\mathbf{v}’$ is
\begin{align*}
\mathbf{v}+\mathbf{v}’&=(a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k)+(b_1\mathbf{v}_1+b_2\mathbf{v}_2+\cdots+ b_k\mathbf{v}_k)\\
&=(a_1+b_1)\mathbf{v}_1+(a_2+b_2)\mathbf{v}_2+\cdots+(a_k+b_k)\mathbf{v}_k.
\end{align*}
The the sum of the coefficients of the above linear combination is
\begin{align*}
&(a_1+b_1)+(a_2+b_2)+\cdots+(a_k+b_k)\\
&=(a_1+a_2+\cdots+a_k)+(b_1+b_2+\cdots+b_k) \stackrel{(*)}{=} 0+0=0.
\end{align*}
It follows that the sum $\mathbf{v}+\mathbf{v}’$ is in $W$, and hence condition 2 is met.

Finally, let us check condition 3. Let
\[\mathbf{v}=a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k\] be an arbitrary vector in $W$ and let $c\in K$.
Since $\mathbf{v}\in W$, we have
\[a_1+a_2+\cdots+a_k=0.\] Then the scalar product is
\begin{align*}
c\mathbf{v}&=c(a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k)\\
&=ca_1\mathbf{v}_1+ca_2\mathbf{v}_2+\cdots+ ca_k\mathbf{v}_k.
\end{align*}
The sum of the coefficients of the linear combination is
\begin{align*}
ca_1+ca_2+\cdots+ ca_k=c(a_1+a_2+\cdots+a_k)=a0=0.
\end{align*}
Hence $c\mathbf{v}\in V$, and condition 3 is met.

Therefore by the subspace criteria, we conclude that $W$ is a subspace of $V$.

Proof 2. (Span)

Consider an arbitrary vector in $W$:
\[a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k \text{ with } a_1+a_2+\cdots+a_k=0.\] Substituting the relation $a_k=-(a_1+a_2+\cdots+a_{k-1})$, we obtain
\begin{align*}
&a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+a_{k-1}\mathbf{v}_{k-1}+ a_k\mathbf{v}_k\\
&=a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots +a_{k-1}\mathbf{v}_{k-1}-(a_1+a_2+\cdots+a_{k-1})\mathbf{v}_k\\
&=a_1(\mathbf{v}_1-\mathbf{v}_k)+a_2(\mathbf{v}_2-\mathbf{v}_k)+\cdots+a_{k-1}(\mathbf{v}_{k-1}-\mathbf{v}_{k-1}).
\end{align*}
This computation yields that every vector in $W$ is a linear combination of vectors in
\[S:=\{\mathbf{v}_1-\mathbf{v}_k, \mathbf{v}_2-\mathbf{v}_k,\dots, \mathbf{v}_{k-1}-\mathbf{v}_{k-1}\}.\] That is, we have $W\subset \Span(S)$.

On the other hand, let
\[\mathbf{v}=c_1(\mathbf{v}_1-\mathbf{v}_k)+c_2(\mathbf{v}_2-\mathbf{v}_k)+\cdots+c_{k-1}(\mathbf{v}_{k-1}-\mathbf{v}_{k-1})\] be an arbitrary vector in $\Span(S)$.
Then we have
\begin{align*}
\mathbf{v}&=c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots +c_{k-1}\mathbf{v}_{k-1}-(c_1+c_2+\cdots+c_{k-1})\mathbf{v}_k.
\end{align*}
This is a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$, and the sum of the coefficients is
\[c_1+c_2+\cdots +c_{k-1}-(c_1+c_2+\cdots+c_{k-1})=0.\] Therefore $\mathbf{v}\in W$. Thus we also have $\Span(S)\subset W$.

Putting together these inclusion yields that $W=\Span(S)$.
As the span is always a subspace, we conclude that $W$ is a subspace of $V$.

The post The Subspace of Linear Combinations whose Sums of Coefficients are zero first appeared on Problems in Mathematics.

(a) Let $S$ be the subset of $\R^4$ consisting of vectors $\begin{bmatrix}
x \\
y \\
z \\
w
\end{bmatrix}$ satisfying
\[2x+4y+3z+7w+1=0.\] Determine whether $S$ is a subspace of $\R^4$. If so prove it. If not, explain why it is not a subspace.

(b) Let $S$ be the subset of $\R^4$ consisting of vectors $\begin{bmatrix}
x \\
y \\
z \\
w
\end{bmatrix}$ satisfying
\[2x+4y+3z+7w=0.\] Determine whether $S$ is a subspace of $\R^4$. If so prove it. If not, explain why it is not a subspace.

(These two problems look similar but note that the equations are different.)

(The Ohio State University, Linear Algebra Final Exam Problem)
 

Solution.

(a) $2x+4y+3z+7w+1=0$

We claim that $S$ is not a subspace of $\R^4$.
If $S$ is a subspace of $\R^4$, then the zero vector $\mathbf{0}=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}$ in $\R^4$ must lie in $S$.

However, the zero vector $\mathbf{0}$ does not satisfy the equation
\[2x+4y+3z+7w+1=0.\]

So $\mathbf{0} \not \in S$, and we conclude that $S$ is not subspace of $\R^4$.

(b) $2x+4y+3z+7w=0$

In a set theoretical notation, we have
\[S=\left\{\, \begin{bmatrix}
x \\
y \\
z \\
w
\end{bmatrix}\in \R^4 \quad \middle| \quad 2x+4y+3z+7w=0 \,\right\}.\]

Let $A$ be the $1\times 4$ matrix defined by
\[A=\begin{bmatrix}
2 & 4 & 3 & 7
\end{bmatrix}.\] Then the equation $2x+4y+3z+7w=0$ can be written as
\[A\begin{bmatrix}
x \\
y \\
z \\
w
\end{bmatrix}=0.\]

So we have
\begin{align*}
S&=\left\{\, \begin{bmatrix}
x \\
y \\
z \\
w
\end{bmatrix}\in \R^4 \quad \middle| \quad A\begin{bmatrix}
x \\
y \\
z \\
w
\end{bmatrix}=0 \,\right\}\\
&=\calN(A),
\end{align*}
the null space of $A$.

Recall that the null space of a matrix is always a subspace.
Hence the subset $S$ is a subspace of $\R^4$ as it is the null space of the matrix $A$.

Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University)

This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568).

The other problems can be found from the links below.

1. Find All the Eigenvalues of 4 by 4 Matrix
2. Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue
3. Diagonalize a 2 by 2 Matrix if Diagonalizable
4. Find an Orthonormal Basis of the Range of a Linear Transformation
5. The Product of Two Nonsingular Matrices is Nonsingular
6. Determine Whether Given Subsets in ℝ4 R 4 are Subspaces or Not (This page)
7. Find a Basis of the Vector Space of Polynomials of Degree 2 or Less Among Given Polynomials
8. Find Values of $a , b , c$ such that the Given Matrix is Diagonalizable
9. Idempotent Matrix and its Eigenvalues
10. Diagonalize the 3 by 3 Matrix Whose Entries are All One
11. Given the Characteristic Polynomial, Find the Rank of the Matrix
12. Compute $A^{10}\mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$
13. Determine Whether There Exists a Nonsingular Matrix Satisfying $A^4=ABA^2+2A^3$

The post Determine Whether Given Subsets in $\R^4$ are Subspaces or Not first appeared on Problems in Mathematics.

Let $V$ be a vector space over a field $K$.
If $W_1$ and $W_2$ are subspaces of $V$, then prove that the subset
\[W_1+W_2:=\{\mathbf{x}+\mathbf{y} \mid \mathbf{x}\in W_1, \mathbf{y}\in W_2\}\] is a subspace of the vector space $V$.

Proof.

We prove the following subspace criteria:

1. The zero vector $\mathbf{0}$ of $V$ is in $W_1+W_2$.
2. For any $\mathbf{u}, \mathbf{v}\in W_1+W_2$, we have $\mathbf{u}+\mathbf{v}\in W_1+W_2$.
3. For any $\mathbf{v}\in W_1+W_2$ and $r\in K$, we have $r\mathbf{v}\in W_1+W_2$.

Since $W_1$ and $W_2$ are subspaces of $V$, the zero vector $\mathbf{0}$ of $V$ is in both $W_1$ and $W_2$.
Thus we have
\[\mathbf{0}=\mathbf{0}+\mathbf{0}\in W_1+W_2.\] So condition 1 is met.

Next, let $\mathbf{u}, \mathbf{v}\in W_1+W_2$.
Since $\mathbf{u}\in W_1+W_2$, we can write
\[\mathbf{u}=\mathbf{x}+\mathbf{y}\] for some $\mathbf{x}\in W_1$ and $\mathbf{y}\in W_2$.
Similarly, we write
\[\mathbf{v}=\mathbf{x}’+\mathbf{y}’\] for some $\mathbf{x}’\in W_1$ and $\mathbf{y}’\in W_2$.

Then we have
\begin{align*}
\mathbf{u}+\mathbf{v}&=(\mathbf{x}+\mathbf{y})+(\mathbf{x}’+\mathbf{y}’)\\
&=(\mathbf{x}+\mathbf{x}’)+(\mathbf{y}+\mathbf{y}’).
\end{align*}
Since $\mathbf{x}$ and $\mathbf{x}’$ are both in the vector space $W_1$, their sum $\mathbf{x}+\mathbf{x}’$ is also in $W_1$.
Similarly we have $\mathbf{y}+\mathbf{y}’\in W_2$ since $\mathbf{y}, \mathbf{y}’\in W_2$.

Thus from the expression above, we see that
\[\mathbf{u}+\mathbf{v}\in W_1+W_2,\] hence condition 2 is met.

Finally, let $\mathbf{v}\in W_1+W_2$ and $r\in K$.
Then there exist $\mathbf{x}\in W_1$ and $\mathbf{y}\in W_2$ such that
\[\mathbf{v}=\mathbf{x}+\mathbf{y}.\] Since $W_1$ is a subspace, it is closed under scalar multiplication. Hence we have $r\mathbf{x}\in W_1$.
Similarly, we have $r\mathbf{y}\in W_2$.

It follows from this observation that
\begin{align*}
r\mathbf{v}&=r(\mathbf{x}+\mathbf{y})\\
&=r\mathbf{x}+r\mathbf{y}\in W_1+W_2,
\end{align*}
and thus condition 3 is met.

Therefore, by the subspace criteria $W_1+W_2$ is a subspace of $V$.

Related Question.

For a proof, see the post “Dimension of the sum of two subspaces“.

The post The Sum of Subspaces is a Subspace of a Vector Space first appeared on Problems in Mathematics.