Find the Dimension of the Subspace of Vectors Perpendicular to Given Vectors
Problem 578
Let $V$ be a subset of $\R^4$ consisting of vectors that are perpendicular to vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$, where
\[\mathbf{a}=\begin{bmatrix}
1 \\
0 \\
1 \\
0
\end{bmatrix}, \quad \mathbf{b}=\begin{bmatrix}
1 \\
1 \\
0 \\
0
\end{bmatrix}, \quad \mathbf{c}=\begin{bmatrix}
0 \\
1 \\
-1 \\
0
\end{bmatrix}.\]
Namely,
\[V=\{\mathbf{x}\in \R^4 \mid \mathbf{a}^{\trans}\mathbf{x}=0, \mathbf{b}^{\trans}\mathbf{x}=0, \text{ and } \mathbf{c}^{\trans}\mathbf{x}=0\}.\]
(a) Prove that $V$ is a subspace of $\R^4$.
(b) Find a basis of $V$.
(c) Determine the dimension of $V$.
Sponsored Links
Contents
Proof.
(a) Prove that $V$ is a subspace of $\R^4$.
Observe that the conditions
\[\mathbf{a}^{\trans}\mathbf{x}=0, \mathbf{b}^{\trans}\mathbf{x}=0, \text{ and } \mathbf{c}^{\trans}\mathbf{x}=0\]
can be combined into the following matrix equation
\[A\mathbf{x}=\mathbf{0},\]
where
\[A=\begin{bmatrix}
1 & 0 & 1 & 0 \\
1 &1 & 0 & 0 \\
0 & 1 & -1 & 0
\end{bmatrix}\]
and $\mathbf{0}$ is the three dimensional zero vector.
Note that the rows of the matrix $A$ are $\mathbf{a}^{\trans}$, $\mathbf{b}^{\trans}$, and $\mathbf{c}^{\trans}$.
It follows that the subset $V$ is the null space $\calN(A)$ of the matrix $A$.
Being the null space, $V=\calN(A)$ is a subspace of $\R^4$.
(See the post “The Null Space (the Kernel) of a Matrix is a Subspace of $\R^n$“.)
(b) Find a basis of $V$.
In the proof of Part (a), we saw that $V=\calN(A)$.
To find a basis, we determine the solutions of $A\mathbf{x}=\mathbf{0}$.
Applying elementary row operations to the augmented matrix, we see that
\begin{align*}
\left[\begin{array}{rrrr|r}
1 & 0 & 1 & 0 &0 \\
1 & 1 & 0 & 0 & 0 \\
0 & 1 & -1 & 0 & 0
\end{array}\right]
\xrightarrow{R_2-R_1}
\left[\begin{array}{rrrr|r}
1 & 0 & 1 & 0 &0 \\
0 & 1 & -1 & 0 & 0 \\
0 & 1 & -1 & 0 & 0
\end{array}\right]\\[6pt]
\xrightarrow{R_3-R_2}
\left[\begin{array}{rrrr|r}
1 & 0 & 1 & 0 &0 \\
0 & 1 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array}\right].
\end{align*}
It follows that the general solution is given by
\[x_1=-x_3, x_2=x_3.\]
The vector form solution is
\[\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}=\begin{bmatrix}
-x_3 \\
x_3 \\
x_3 \\
x_4
\end{bmatrix}=x_3\begin{bmatrix}
-1 \\
1 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix}.\]
Hence we have
\begin{align*}
V&=\calN(A)\\
&=\left\{\, \mathbf{x}\in \R^4 \quad \middle|\quad \mathbf{x}=x_3\begin{bmatrix}
-1 \\
1 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix}, \text{ where $x_3, x_4\in \R$} \,\right\}\\[6pt]
&=\Span \left\{\, \begin{bmatrix}
-1 \\
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix} \,\right\}.
\end{align*}
Let $B:=\left\{\, \begin{bmatrix}
-1 \\
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix} \,\right\}$.
Then we just showed that $B$ is a spanning set for $V$.
It is straightforward to see that $B$ is linearly independent.
Hence $B$ is a basis for $V$.
(c) Determine the dimension of $V$.
As the basis $B$ for $V$ that we obtained in Part (b) consists of two vectors, the dimension of the subspace $V$ is $\dim(V)=2$.
Add to solve later
Sponsored Links
More from my site
The Subset Consisting of the Zero Vector is a Subspace and its Dimension is Zero
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 […]
Prove a Given Subset is a Subspace and Find a Basis and Dimension
Let
\[A=\begin{bmatrix}
4 & 1\\
3& 2
\end{bmatrix}\]
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 $\R^2$.
(b) Find a basis for […]- Find a Basis and the Dimension of the Subspace of the 4-Dimensional Vector Space
Let $V$ be the following subspace of the $4$-dimensional vector space $\R^4$.
\[V:=\left\{ \quad\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix} \in \R^4
\quad \middle| \quad
x_1-x_2+x_3-x_4=0 \quad\right\}.\]
Find a basis of the subspace $V$ […]
- Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis
Let $P_3$ be the vector space over $\R$ of all degree three or less polynomial with real number coefficient.
Let $W$ be the following subset of $P_3$.
\[W=\{p(x) \in P_3 \mid p'(-1)=0 \text{ and } p^{\prime\prime}(1)=0\}.\]
Here $p'(x)$ is the first derivative of $p(x)$ and […]
- True or False. The Intersection of Bases is a Basis of the Intersection of Subspaces
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 $\R^n$.
If $B_1$ and $B_2$ are bases for $W_1$ and $W_2$, respectively, then $B_1\cap B_2$ is a […]
Quiz 6. Determine Vectors in Null Space, Range / Find a Basis of Null Space
(a) Let $A=\begin{bmatrix}
1 & 2 & 1 \\
3 &6 &4
\end{bmatrix}$ and let
\[\mathbf{a}=\begin{bmatrix}
-3 \\
1 \\
1
\end{bmatrix}, \qquad \mathbf{b}=\begin{bmatrix}
-2 \\
1 \\
0
\end{bmatrix}, \qquad \mathbf{c}=\begin{bmatrix}
1 \\
1 […]- Row Equivalent Matrix, Bases for the Null Space, Range, and Row Space of a Matrix
Let \[A=\begin{bmatrix}
1 & 1 & 2 \\
2 &2 &4 \\
2 & 3 & 5
\end{bmatrix}.\]
(a) Find a matrix $B$ in reduced row echelon form such that $B$ is row equivalent to the matrix $A$.
(b) Find a basis for the null space of $A$.
(c) Find a basis for the range of $A$ that […]
- Rank and Nullity of a Matrix, Nullity of Transpose
Let $A$ be an $m\times n$ matrix. The nullspace of $A$ is denoted by $\calN(A)$.
The dimension of the nullspace of $A$ is called the nullity of $A$.
Prove the followings.
(a) $\calN(A)=\calN(A^{\trans}A)$.
(b) $\rk(A)=\rk(A^{\trans}A)$.
Hint.
For part (b), […]
