Question

Exercise 2 Find bases and dimensions for the four subspaces associated with A and B A= (1 2 4 2 4 8 B= 1 2 4 2 5 8

Answer 1

The 4 subspaces associated with a given matrix A are:

• Row space-Vector space generated by rows of the matrix
• Column space-Vector space generated by columns of the matrix
• Null space (kernel)-Set of vectors X such that AX=O
• Left null space-Set of vectors Y such that
  Y? 4 = 0
  ..

A basis for a vector space is a set of linearly independent vectors of the space, that can generate the space.

(For this reason, the zero vector cannot be in the basis of any vector space, as it can be written as a linear combination of all vectors, irrespective of type and number, simply by taking all scalars in the linear combination as 0).

(Linearly independent set of vectors: A set of vectors such that not a single vector of the set can be expressed as a linear combination of the other vectors of that set.)

Let us consider the matrix

$A=\begin{bmatrix} 1 & 2 & 4\\ 2& 4 & 8 \end{bmatrix}$.

The rows of this matrix are
1 2 4.2 48
. So, the row space will be the vector space generated by these vectors. Now,
24=124
. So,
2 8
is a scalar multiple of , which means that the space generated by
1 2 4.2 48
can also be generated by alone. So,
{124
is a basis for the row space of A, a basis with 1 element.

A basis for row space of A is , and so its dimension is 1.

The columns of matrix A are $\begin{bmatrix} 1\\ 2 \end{bmatrix},\begin{bmatrix} 2\\ 4 \end{bmatrix},\begin{bmatrix} 4\\ 8 \end{bmatrix}$ . Now, $\begin{bmatrix} 2\\ 4 \end{bmatrix}=2\begin{bmatrix} 1\\ 2 \end{bmatrix} ,\begin{bmatrix} 4\\ 8 \end{bmatrix}=4\begin{bmatrix} 1\\ 2 \end{bmatrix}$ , which means $\begin{bmatrix} 2\\ 4 \end{bmatrix},\begin{bmatrix} 4\\ 8 \end{bmatrix}$ are scalar multiples of $\begin{bmatrix} 1\\ 2 \end{bmatrix}$ . So, the column space of A, or the vector space generated by columns of A, can be generated by $\begin{bmatrix} 1\\ 2 \end{bmatrix}$ alone. So, $\left\{\begin{bmatrix} 1\\ 2 \end{bmatrix}\right\}$ is a basis for column space of A, and has 1 element.

A basis for column space of A is $\left\{\begin{bmatrix} 1\\ 2 \end{bmatrix}\right\}$ and so its dimension is 1.

Let us consider the vector

$X=\begin{bmatrix} a\\b\\c\end{bmatrix}$ such that AX=O

$\\=>\begin{bmatrix} 1 & 2 &4 \\ 2 & 4& 8 \end{bmatrix} \begin{bmatrix} a\\b\\c\end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix} \\=>\begin{bmatrix} a+ 2b+4c\\ 2a+4b+8c \end{bmatrix} =\begin{bmatrix} 0\\ 0 \end{bmatrix} \\=>a=-2b-4c \\=>\begin{bmatrix} a\\b\\c\end{bmatrix}= \begin{bmatrix} -2b-4c\\b\\c\end{bmatrix}=b \begin{bmatrix} -2\\1\\0\end{bmatrix}+c \begin{bmatrix} -4\\0\\1\end{bmatrix}$

So, any vector in the null space of A can be written as a linear combination of , neither of which can be written as a scalar multiple of the other. So, they are linearly independent vectors that can generate null space of A.

-2 1 0 0 1

A basis for null space of A is
-2 -4 0 1 0 1
, and dimension is 2.

Let us consider the vector

$Y=\begin{bmatrix} p\\q\end{bmatrix}$ such that $Y^TA=O \\=>\begin{bmatrix} p&q\end{bmatrix} \begin{bmatrix} 1 &2 & 4\\ 2& 4& 8 \end{bmatrix}=\begin{bmatrix} 0& 0& 0 \end{bmatrix} \\=>p+2q=0,\ 2p+4q=0,\ 4p+8q=0 \\=>p=-2q \\=>\begin{bmatrix} p\\q\end{bmatrix}= \begin{bmatrix} -2q\\q\end{bmatrix} =q\begin{bmatrix} -2\\1\end{bmatrix}$

So, any vector of the left null space of A can be written as a scalar multiple of $\begin{bmatrix} -2\\1\end{bmatrix}$ .

A basis for the left null space of A is $\left \{\begin{bmatrix} -2\\1\end{bmatrix} \right \}$ and has dimension 1.

Let us consider the matrix

$B=\begin{bmatrix} 1 & 2 & 4\\ 2& 5 & 8 \end{bmatrix}$

The rows of this matrix are . Since neither of these vectors can be represented as a scalar multiple of the other, they are linearly independent. So, is a basis for the row space of B, containing 2 elements.

A basis for row space of B is , and its dimension is 2.

The columns of B are $\begin{bmatrix} 1\\ 2 \end{bmatrix},\begin{bmatrix} 2\\ 5 \end{bmatrix},\begin{bmatrix} 4\\ 8 \end{bmatrix}$ , and $\begin{bmatrix} 4\\ 8 \end{bmatrix}=4\begin{bmatrix} 1\\ 2 \end{bmatrix}$ . So, the vectors $\begin{bmatrix} 1\\ 2 \end{bmatrix},\begin{bmatrix} 2\\ 5 \end{bmatrix}$ are enough to generate the column space of B. Since neither of these 2 vectors can be written as a scalar multiple of the other, these vectors are linearly independent, and they generate the column space of B.

A basis for column space of B is $\left\{ \begin{bmatrix} 1\\ 2 \end{bmatrix},\begin{bmatrix} 2\\ 5 \end{bmatrix}\right\}$ and its dimension is 2.

Let us consider the vector

$X=\begin{bmatrix} a\\b\\c\end{bmatrix}$ such that BX=O

$\\=>\begin{bmatrix} 1 & 2 & 4\\ 2& 5 & 8 \end{bmatrix} \begin{bmatrix} a\\b\\c\end{bmatrix} =\begin{bmatrix} 0\\ 0 \end{bmatrix} \\=>a+2b+4c=0,\ \ 2a+5b+8c=0 \\=>2a+4b+8c=0,\ \ 2a+5b+8c=0 \\=>2a+5b+8c-(2a+4b+8c)=0 \\=>b=0 \\=>a+4c=0=>a=-4c \\=>\begin{bmatrix} a\\b\\c\end{bmatrix}= \begin{bmatrix} -4c\\0\\c\end{bmatrix}=c \begin{bmatrix} -4\\0\\1\end{bmatrix}$

So, any element of the null space of B can be written as a scalar multiple of $\begin{bmatrix} -4\\0\\1\end{bmatrix}$ .

A basis for the null space of B is $\left\{ \begin{bmatrix} -4\\0\\1\end{bmatrix}\right\}$ and its dimension is 1.

Let us consider the vector $Y=\begin{bmatrix} p\\q\end{bmatrix}$ such that $Y^TB=O$

$\\=>\begin{bmatrix} p&q\end{bmatrix} \begin{bmatrix} 1 & 2 & 4\\ 2& 5 & 8 \end{bmatrix}=[0\ \ 0\ \ 0] \\=>p+2q=0, 2p+5q=0,4p+8q=0 \\=>p=-2q,\ p=-\frac52q \\=>p=q=0 \\=>\begin{bmatrix} p\\q\end{bmatrix}= \begin{bmatrix} 0\\0\end{bmatrix}$

So, any element of the left null space of B is a scalar multiple of $\begin{bmatrix} 0\\0\end{bmatrix}$ , which will all ultimately be $\begin{bmatrix} 0\\0\end{bmatrix}$ .

So, the only element of the left null space of B is the zero element $\begin{bmatrix} 0\\0\end{bmatrix}$ . Since the zero element cannot be in the basis, the basis of the left null space is the empty set.

The only basis for the left null space of B is the empty set $\phi$ , and so its dimension is 0.