Two Subspaces Intersecting Trivially, and the Direct Sum of Vector Spaces.

Problems and solutions in Linear Algebra

Problem 61

Let $V$ and $W$ be subspaces of $\R^n$ such that $V \cap W =\{\mathbf{0}\}$ and $\dim(V)+\dim(W)=n$.

(a) If $\mathbf{v}+\mathbf{w}=\mathbf{0}$, where $\mathbf{v}\in V$ and $\mathbf{w}\in W$, then show that $\mathbf{v}=\mathbf{0}$ and $\mathbf{w}=\mathbf{0}$.

(b) If $B_1$ is a basis for the subspace $V$ and $B_2$ is a basis for the subspace $W$, then show that the union $B_1\cup B_2$ is a basis for $R^n$.

(c) If $\mathbf{x}$ is in $\R^n$, then show that $\mathbf{x}$ can be written in the form $\mathbf{x}=\mathbf{v}+\mathbf{w}$, where $\mathbf{v}\in V$ and $\mathbf{w} \in W$.

(d) Show that the representation obtained in part (c) is unique.

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Hint.

Try to show that the vector $\mathbf{v}$ is in both $V$ and $W$, hence in $V \cap W$.

Proof.

(a) Show that $\mathbf{v}=\mathbf{0}$ and $\mathbf{w}=\mathbf{0}

 Suppose that we have $\mathbf{v}+\mathbf{w}=\mathbf{0}$, where $\mathbf{v}\in V$ and $\mathbf{w}\in W$. Then we have $\mathbf{v}=-\mathbf{w}$. Note that the left hand side is in $V$ and the right hand side is in $W$. Thus the vector $\mathbf{v}$ is in both $V$ and $W$, that is $\mathbf{v} \in V \cap W=\{\mathbf{0}\}$. Hence $\mathbf{v}=\mathbf{0}$, and it follows that $\mathbf{w}$ is also the zero vector.

 

(b) Show that the union $B_1\cup B_2$ is a basis for $R^n$

Let $B_1=\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$ and $B_2=\{\mathbf{w}_1, \mathbf{w}_2, \cdots, \mathbf{w}_l\}$ be bases for $V$ and $W$, respectively.
Note that $\dim(V)=k$, $\dim(W)=l$, and $k+l=n$.

We claim that the set $B_1\cup B_2$ is linearly independent.
Suppose that we have a linear combination of vectors in $B_1\cup B_2$
\[c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots+c_k \mathbf{v}_k+d_1 \mathbf{w}_1+d_2 \mathbf{w}_2+\cdots+d_l \mathbf{w}_l=\mathbf{0}. \tag{*}\] Then observe that the vector $\mathbf{v}:=c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots+c_k \mathbf{v}_k$ is in $V$ and the vector $\mathbf{w}:=d_1 \mathbf{w}_1+d_2 \mathbf{w}_2+\cdots+d_l \mathbf{w}_l$ is in $W$. Since their sum $\mathbf{v}+\mathbf{w}=\mathbf{0}$, by part (a), we must have $\mathbf{v}=\mathbf{0}$ and $\mathbf{w}=\mathbf{0}$.

Since $B_1$ is a basis for $V$, the vectors $\mathbf{v}_i$ are linearly independent. Thus
\[\mathbf{0}=\mathbf{v}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots+c_k \mathbf{v}_k\] implies that the coefficients $c_i$ are all zero.

Similarly, since $B_2$ is a basis for $W$,
\[\mathbf{0}=\mathbf{w}=d_1 \mathbf{w}_1+d_2 \mathbf{w}_2+\cdots+d_l \mathbf{w}_l\] implies that we have $d_i=0$.

Therefore the coefficient of the linear combination (*) is all zero. Thus the vectors $\mathbf{v_1},\dots, \mathbf{v_k}, \mathbf{w}_1, \dots, \mathbf{w}_l$ are linearly independent.
Since $B_1 \cup B_2$ consists of $n$ linearly independent vectors in the $n$ dimensional vector space $\R^n$, they must be a basis.

(c) Show that $\mathbf{x}$ can be written in the form $\mathbf{x}=\mathbf{v}+\mathbf{w}$

For any vector $\mathbf{x} \in \R^n$, since $B_1 \cup B_2$ is a basis we have
\[\mathbf{x}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots+c_k \mathbf{v}_k+d_1 \mathbf{w}_1+d_2 \mathbf{w}_2+\cdots+d_l \mathbf{w}_l\] for some $c_1, \dots, c_k,d_1,\dots, d_l \in \R$.

Let
\[\mathbf{v}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots+c_k \mathbf{v}_k\] and
\[\mathbf{w}=d_1 \mathbf{w}_1+d_2 \mathbf{w}_2+\cdots+d_l \mathbf{w}_l.\] Then we have $\mathbf{v} \in V$, $\mathbf{w} \in W$, and $\mathbf{x}=\mathbf{v}+\mathbf{w}$ as required.

(d) Show that the representation obtained in part (c) is unique.

Suppose that we have two representations for $\mathbf{x}$, that is, we have $\mathbf{x}=\mathbf{v}’+\mathbf{w}’$ in addition to the $\mathbf{x}=\mathbf{v}+\mathbf{w}$, where $\mathbf{v}, \mathbf{v}’\in V$ and $\mathbf{w}, \mathbf{w}’ \in W$.

Then we have $\mathbf{v}+\mathbf{w}=\mathbf{x}=\mathbf{v}’+\mathbf{w}’$, or equivalently,
\[\mathbf{v}-\mathbf{v}’=\mathbf{w}’-\mathbf{w}.\]

Observe that the left hand side is in $V$ and the right hand side is in $W$ thus we have
\[\mathbf{v}-\mathbf{v}’, \mathbf{w}’-\mathbf{w} \in V \cap W=\{\mathbf{0}\}.\] Thus we have $\mathbf{v}=\mathbf{v}’$ and $\mathbf{w}=\mathbf{w}’$ and we conclude that the representations are actually identical, hence we proved the uniqueness.


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