Rank of a matrix

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April 1st 2010, 05:17 PM
mybrohshi5

Rank of a matrix

Find the rank of the matrix

A = $\begin{bmatrix}0&6\\0&-2\\0&5\end{bmatrix}$

I know that rank is just the number of pivots a matrix has when in reduced row echelon form, but this one is confusing to me.

RREF A = $\begin{bmatrix}0&1\\0&0\\0&0\end{bmatrix}$

I thought this would have a rank of 0 because there are no pivots but i was wrong and it has a rank of 1.

Why is this?

Thanks for any help :)
April 1st 2010, 05:32 PM
harish21

Quote:

Originally Posted by mybrohshi5

Find the rank of the matrix

A = $\begin{bmatrix}0&6\\0&-2\\0&5\end{bmatrix}$

I know that rank is just the number of pivots a matrix has when in reduced row echelon form, but this one is confusing to me.

RREF A = $\begin{bmatrix}0&1\\0&0\\0&0\end{bmatrix}$

I thought this would have a rank of 0 because there are no pivots but i was wrong and it has a rank of 1.

Why is this?

Thanks for any help :)

The matrix RREF A has one "non-zero" row. that means the matrix A has one independent row vector. So its rank is 1.
April 1st 2010, 05:37 PM
mybrohshi5

So finding rank is just the number of rows that has at least one non-zero entry in it?
April 1st 2010, 05:48 PM
harish21

Quote:

Originally Posted by mybrohshi5

So finding rank is just the number of rows that has at least one non-zero entry in it?

Yes, this is the way that I learnt to find the rank of a matrix. I would also suggest referring to linear independence

Actually, the NUMBER of rows (in the RREF matrix) that are NON-ZERO is the RANK of the matrix. Here your matrix has one $non-zero$ row, your rank is 1.
April 1st 2010, 05:50 PM
mybrohshi5

Thanks that clears things up for me :)