Dimension of 4x4 Matrix: Find Basis Vectors

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Homework Statement


A= \begin{bmatrix}1 & 1 & 1 & 1 \\ 2 & 1 & 0 & -1 \\ 3 & 4 & 5 & 6 \\ -1 &2 &1&0 \end{bmatrix}Determine the dimension of A and give a set of basis vectors for A.


Homework Equations


Dimension of matrix, ref form of matrix.


The Attempt at a Solution


I reduced the matrix to row echelon form and then the dimension = rank of matrix. Is that correct? I am quite confused about what dimension means. In a 4x4 matrix, maybe dimension is 16? or is it the number of non-zero matrix elements?
 
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Easy way to remember it is:

# of columns - # of non zero rows in rref
 
So, the dimension of A is the rank of the matrix A?
 
Yes. In fact, I would consider the term "dimension of a matrix" very strange. The rank of a matrix is the dimension of the image of the matrix.
 
I think by dimension you mean "nullity" cause our lecturer also used "dimension of a matrix" which was confusing when studying from other sources.

Try plugging your matrix in wolfram and ask for nullity.

In the output it states a "dimension" which is always exactly what I always needed. So maybe that's what it means.
 
sharks said:

Homework Statement


A= \begin{bmatrix}1 & 1 & 1 & 1 \\ 2 & 1 & 0 & -1 \\ 3 & 4 & 5 & 6 \\ -1 &2 &1&0 \end{bmatrix}Determine the dimension of A and give a set of basis vectors for A.


Homework Equations


Dimension of matrix, ref form of matrix.


The Attempt at a Solution


I reduced the matrix to row echelon form and then the dimension = rank of matrix. Is that correct? I am quite confused about what dimension means. In a 4x4 matrix, maybe dimension is 16? or is it the number of non-zero matrix elements?

How does your textbook or lecturer or course notes define the term "dimension of a matrix"?

RGV
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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