Prove a Given Subset is a Subspace and Find a Basis and Dimension
Problem 270
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 $V$ and determine the dimension of $V$.
Sponsored Links
Contents
Proof.
(a) $V$ is a subspace of $\R^2$
Note that the equality $A\mathbf{x}=5\mathbf{x}$ can be written as
\[(A-5I)\mathbf{x}=\mathbf{0},\]
where $I$ is the $2 \times 2$ matrix.
If we define $B=A-5I$, then the subset $V$ becomes
\begin{align*}
V&=\{\mathbf{x}\in \R^2 \mid B\mathbf{x}=\mathbf{0}\}\\
&=\calN(B),
\end{align*}
the null space of the matrix $B$.
Since any null space is a vector space, this shows that $V$ is a subspace of $\R^2$.
(b) A basis for $V$ and the dimension of $V$
From part (a), we have obtained that
\[V=\calN(B),\]
where
\[B=A-5I=\begin{bmatrix}
-1 & 1\\
3& -3
\end{bmatrix}.\]
Thus, $V$ is a set of solutions of the system $B\mathbf{x}=\mathbf{0}$.
The augmented matrix for this system is reduced as follows.
\[\left[\begin{array}{rr|r}
-1 & 1 & 0 \\
3 &-3 &0
\end{array} \right]
\xrightarrow{R_2+3R_1}
\left[\begin{array}{rr|r}
-1 & 1 & 0 \\
0 & 0 & 0
\end{array} \right]
\xrightarrow{-R_1}
\left[\begin{array}{rr|r}
1 & -1 & 0 \\
0 & 0 & 0
\end{array} \right].
\]
From this we see that the general solution $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}$ satisfies
\[x_1=x_2,\]
and thus the general solution is
\[\mathbf{x}=\begin{bmatrix}
x_2 \\
x_2
\end{bmatrix}=x_2\begin{bmatrix}
1 \\
1
\end{bmatrix}.\]
Therefore we have
\begin{align*}
V&=\left \{ \mathbf{x} \in \R^2 \quad \middle| \quad \mathbf{x}=x_2\begin{bmatrix}
1 \\
1
\end{bmatrix} \text{ for any } x_2 \in \R \right \}\\
&=\Span\left\{ \begin{bmatrix}
1 \\
1
\end{bmatrix} \right \}.
\end{align*}
Thus the set
\[\left\{ \begin{bmatrix}
1 \\
1
\end{bmatrix} \right \}\]
is a spanning set of $V$ and it is linearly independent as it consists of only a single vector.
Therefore
\[\left\{ \begin{bmatrix}
1 \\
1
\end{bmatrix} \right \}\]
is a basis for the subspace $V$, and thus the dimension of $V$ is $1$.
(Recall that the dimension of a vector space is the number of vectors in a basis for the vector space.)
Add to solve later
Sponsored Links
More from my site
Dimension of the Sum of Two Subspaces
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 […]- Determine Null Spaces of Two Matrices
Let
\[A=\begin{bmatrix}
1 & 2 & 2 \\
2 &3 &2 \\
-1 & -3 & -4
\end{bmatrix} \text{ and }
B=\begin{bmatrix}
1 & 2 & 2 \\
2 &3 &2 \\
5 & 3 & 3
\end{bmatrix}.\]
Determine the null spaces of matrices $A$ and $B$.
Proof.
The null space of the […]
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), […]- A Basis for the Vector Space of Polynomials of Degree Two or Less and Coordinate Vectors
Show that the set
\[S=\{1, 1-x, 3+4x+x^2\}\]
is a basis of the vector space $P_2$ of all polynomials of degree $2$ or less.
Proof.
We know that the set $B=\{1, x, x^2\}$ is a basis for the vector space $P_2$.
With respect to this basis $B$, the coordinate […]
- Find a Basis and Determine the Dimension of a Subspace of All Polynomials of Degree $n$ or Less
Let $P_n(\R)$ be the vector space over $\R$ consisting of all degree $n$ or less real coefficient polynomials. Let
\[U=\{ p(x) \in P_n(\R) \mid p(1)=0\}\]
be a subspace of $P_n(\R)$.
Find a basis for $U$ and determine the dimension of $U$.
Solution.
[…]
- Find a Basis For the Null Space of a Given $2\times 3$ Matrix
Let
\[A=\begin{bmatrix}
1 & 1 & 0 \\
1 &1 &0
\end{bmatrix}\]
be a matrix.
Find a basis of the null space of the matrix $A$.
(Remark: a null space is also called a kernel.)
Solution.
The null space $\calN(A)$ of the matrix $A$ is by […]
- The Null Space (the Kernel) of a Matrix is a Subspace of $\R^n$
Let $A$ be an $m \times n$ real matrix. Then the null space $\calN(A)$ of $A$ is defined by
\[ \calN(A)=\{ \mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}_m\}.\]
That is, the null space is the set of solutions to the homogeneous system $A\mathbf{x}=\mathbf{0}_m$.
Prove that the […]
