- #1

ND3G

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**Find a basis for and calculate the dimension of nullA:**

A = [1 2 7]T, [1 1 2]T, [-2 0 6]T, [0 1 -10]T, [4 -5 -7]T

A = [1 2 7]T, [1 1 2]T, [-2 0 6]T, [0 1 -10]T, [4 -5 -7]T

Like most algebra texts mine has pages and pages of proofs with hardly a single example tying it together.

Here is what I think I know:

If the determinant does not equal 0 than the vectors span R^n and are linearly independent. Therefore, they can makeup a basis for A.

In this case since it is not a square matrix I cannot use determinant to find the solution so I opted to find whether it not it spans R^n such as:

X1 + 2X2 + 7X3 = 0

X1 + X2 + 2X3 = 0

-2X1 + 6X3 = 0

X2 - 10X3 = 0

4X1 -5X2 -7X3 = 0

The above has a non-trivial solution so the vectors are linearly dependent and do not form a basis.

Now I could add a remove vectors until I find a set of vectors with only the trivial solution but it is a pretty time intensive process, especially for a large matrix.

Is there a faster way?