Problem of the week - Column space of a matrix

In Linear Algebra the most important subspaces are tied to matrices. One of these subspaces is the Column Space, which consists of all linear combinations of the columns of a matrix. This subspace, spanned by the columns of a matrix, is crucial in Linear Algebra and it is related to 4 of the most important subjects: Matrices, System of Linear Equations, Vector Spaces and Linear Transformations.

Two interesting results that link these subjects are:

• The system of linear equations associated with the matrix A: Ax=b, is solvable, if and only if b is in the column space of A, which means that b can be expressed as a linear combination of the columns of A.

• The column space of a matrix is the image of the corresponding matrix transformation, and as the rank of a matrix is the same as the dimension of the image, we can conclude that the dimension of the column space of A is equal to the rank of A.

This problem of the week is about this subspace, but you should also apply the concept of symmetric matrix.

Construct, if possible, a matrix A, such that A is symmetric, and its column space is spanned by the vectors (1 -2 5 0) and (-2 0 2 -4).

 

As the matrix is symmetric and the vectors belong to R⁴, A is a 4x4 matrix, and all the columns of the matrix should be a linear combination of the vectors (1 -2 5 0) y (-2 0 2 -4).

Although to get all possible matrices, it would be necessary to express the first two columns as linear combination of the two given vectors, we will assume, as a simplification of the problem, that the two first columns match the given vectors, based on the fact that the second component of the first vector is equal to the first component of the second vector.

As the matrix is symmetric, the first two rows of the matrix must match the given vectors, and a₃₄ must be equal to a₄₃.

┌                ┐
│  1  -2   5   0 │
│ -2   0   2  -4 │
│  5   2   x   w │
│  0  -4   y   z │
└                ┘

The next step is to determine if there exist x, y, w and z such that, when expressing the last two columns as a linear combination of the two first columns, y = w.

We can use the first two rows to get the coefficients when expressing the third and fourth column as a linear combination of the two first ones.

For the third column, the augmented matrix for the system of linear equations related to the linear combination is:

┌             ┐
│  1  -2 |  5 │
│ -2   0 |  2 │
└             ┘

which is transformed to the following row echelon form:

r2 <———> r2 + 2•r1
┌             ┐
│ 1  -2 |   5 │
│ 0  -4 |  12 │
└             ┘

and the solution is

 α = -1
 β = -3

Which means that the third column can be expressed as a linear combination of the first and second column using coefficients -1 and -3 respectively.

Using these coefficients, we can compute the value of x and y, using the third and fourth row.

(-1)5+(-3)2    = x = -11
(-1)0+(-3)(-4) = y =  12

In the same way, we can compute the coefficients for the fourth column.

┌             ┐
│  1  -2 |  0 │
│ -2   0 | -4 │
└             ┘

r2 <———> r2 + 2•r1
┌             ┐
│ 1  -2 |   0 │
│ 0  -4 |  -4 │
└             ┘

where

 α = 2
 β = 1

And again, use the coefficients to compute the value of w and z.

(2)5 + 1(2)  = w = 12
(2)0 + 1(-4) = z = -4

As y = w = 12, it is possible to construct the requested matrix:

    ┌                 ┐
    │  1  -2    5   0 │
    │ -2   0    2  -4 │
A = │  5   2  -11  12 │
    │  0  -4   12  -4 │
    └                 ┘