Dimension of row/ column space

jeffreylze
Messages
43
Reaction score
0

Homework Statement



In the following exercises verify that the row rank is equal to the column rank by explicitly finding the dimensions of the row space and the column space of the given matrix.

A = [1 2 1 ; 2 1 -1]


Homework Equations





The Attempt at a Solution



All i can think of is just row reduce it to row echelon form and then find the rank of the matrix. How do i do it explicitly?
 
Physics news on Phys.org
1. Show that the rows of A are not linear combinations of each other, i.e. one is the multiple of the other.

2. Show that one column is the linear combination of the other 2 columns. Then show that the remaining 2 columns are not a multiple of the other.

then you've explicitly shown that the rank(A) = row rank (A) = column rank (A)
 
hokie1 said:
1. Show that the rows of A are not linear combinations of each other, i.e. one is the multiple of the other.

2. Show that one column is the linear combination of the other 2 columns. Then show that the remaining 2 columns are not a multiple of the other.

then you've explicitly shown that the rank(A) = row rank (A) = column rank (A)

Do you mean by one is NOT the multiple of the other?
 
You are quite correct. I did not proofread before submitting.
 
Okay, thanks. Now that make sense =D
 
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...
Back
Top