Question

linear algebra: basis and dimension

Find all values of a for which {[a^2 0 1], [0 a 2], [1 0 1]} is a basis for R3.

Answer: a cannot be -1, 0, 1.

How do you solve this question? All I know is that the system has to be linearly independent but since the matrix consists of some variables, I don't know how to row reduce it..

Answer 1

|a^2 0 1|
|0 a 2|= 0
|1 0 1|
a^2*a -a =0
a(a^2-1)=0
a=0,a=+-1

Answer 2

got it?

Answer 3

I still don't understand.. how did you get a^2*a -a=0?

Answer 4

for 3 vectors form a basis of R3 they have to be independent. One way to check if vectors are independent is to see if the determinant of the matrix formed by them is equal to 0 or not. If it is 0, it means vectors are dependent.

Answer 5

Oh, but we didn't learn determinants yet. So we're supposed to approach this question using only the knowledge of basis, dimension, span, and subspace.. is there another way to solve this question?

Answer 6

solve the homogeneous system of 3 equations with 3 unknowns

Answer 7

I tried doing that but I don't know how exactly to do it. I did: \[\left[\begin{matrix}a^2 & 0 &1\\ 0 & a & 0 \\ 1 & 2 &1\end{matrix}\right]= \left[\begin{matrix}0 \\ 0\\0\end{matrix}\right]\] (let the right side be an augmented matrix)

Answer 8

But then I don't know what to do from here

Answer 9

You could use row reduction
btw, the original question has the last row as 1 0 1 (not 1 2 1)
if we start with
\[\left[\begin{matrix}a^2 & 0 & 1 \\ 0 & a & 0 \\ 1& 0 &1\end{matrix}\right]\]
multiply the first row by -1/a^2 and add to the 3rd row
\[\left[\begin{matrix}a^2 & 0 & 1 \\ 0 & a & 0 \\ 0& 0 &1-\frac{1}{a^2}\end{matrix}\right]\]
for a full rank matrix we see that a ≠ 0, and 1-1/a^2 ≠ 0
this last implies a^2 ≠ 1 or a≠ 1 and a ≠ -1

Answer 10

Isn't the last row 1 2 1 since the third element of the second vector is a 2 (from [0 a 2])?