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Determine the dimensions of nul a and col a for the matrix shown below.

– $ \begin{bmatrix}
1 & -6 & 9 & 0 & -2\\ 0 & 1 & 2 & -4 &  5\\ 0 & 0 & 0 & 5 &  1\\ 0 & 0 & 0 & 0 &  0 \end{bmatrix}
$

The main objective of this question is to find the null and column space of the given matrix.

This question uses the concept of null space and column space of the matrix. The dimensions of null space and column space are determined by reducing the matrix to a reduced echelon form. The dimension of a null space is determined by the number of variables in the solution, whereas the dimension of its column space is determined by the number of pivots in the matrix’s reduced row-echelon form.

Expert Answer

We have to find the null space and column space of the given matrix. Given that:

\[ \space = \space \begin{bmatrix}
1 & -6 & 9 & 0 & -2\\ 0 & 1 & 2 & -4 &  5\\ 0 & 0 & 0 & 5 &  1\\ 0 & 0 & 0 & 0 &  0 \end{bmatrix} \]

We know that:

\[ \space Ax \space = \space 0 \]

The given matrix is already in reduced echelon form, so:

The dimension of null space of the given matrix is $ 2 $ while the dimension of null space of column $ A $ is  $ 3 $. 

Numerical Answer

The given matrix has a dimension of null space of $ 2 $ and the dimension of column space is $ 3 $. 

Example

Find the null space and column space of the given matrix.

\[ \space = \space \begin{bmatrix}
1 & – 2 & – 5 & 3 & 0\\ -2 & 5 & -2 & -4 &  1 \end{bmatrix} \]

Given that:

\[ \space = \space \begin{bmatrix} 1 & – 2 & – 5 & 3 & 0\\ -2 & 5 & -2 & -4 &  1 \end{bmatrix} \]

We have to find the dimension of null space and column space of the given matrix.

We know that:

\[ \space Ax \space = \space 0 \]

The augmented matrix is:

\[ \space = \space \begin{bmatrix} 1 & – 2 & – 5 & 3 & 0 & 0\\ -2 & 5 & -2 & -4 &  1 & 0 \end{bmatrix} \]

By reducing the given matrix to a reduced echelon form, we get:

\[ \space = \space \begin{bmatrix} 1 & 0 & – 29 & 7 & 2 & 0\\ 0 & 1 & -12 & 2 &  1 & 0 \end{bmatrix} \]

Thus:

\[ \space x \space = \space \begin{bmatrix}
29\\ 12\\ 1\\ 0\\  0 \end{bmatrix}  s \space +   \space \begin{bmatrix} -7 \\ -2\\ 0\\ 1\\  0 \end{bmatrix} t \space + \space  \begin{bmatrix}-2\\ -1\\ 0\\ 0\\  1 \end{bmatrix}    \]

Hence, the dimension of the null space is $ 3 $ and the dimension of the column space is $ 2 $.

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