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I'm confused about this problem. I have four vectors $v_1 = (1,1,1,a), v_2 = (1,2,3,a), v_3= (b,1,0,1), v_4 = (0,b,0,0)$ with $a,b$ real numbers. Determine the maximum and minimum dimension of the space generated by $\{v_1,v_2,v_3,v_4\}\ $

First i put the matrix respect this vector and fount $ A = \begin{bmatrix} 1 & 1 & 1 & a \\ 1 & 2 & 3 & a \\ b & 1 & 0 & 1 \\ 0 & b & 0 & 0 \end{bmatrix}$ = $ \begin{bmatrix} 1 & 0 & -1 & a \\ 0 & 1 & 2 & 0 \\ b & 1 & 0 & 1 \\ 0 & b & 0 & 0 \end{bmatrix}$

but how can I get the maximum and minimum dimension? Some help please.

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Can the dimension ever be $0$? No, because $\{v_1\}$ always spans a $1$ dimensional subspace of $\mathbb{R}^4$.

Can the dimension ever be $1$? No, because $\{v_1,v_2\}$ always spans a $2$ dimensional subspace of $\mathbb{R}^4$.

Can the dimension ever be $2$? Suppose it is, then $v_3=sv_1+tv_2$ and so $1=s+2t$ and $0=s+3t$ giving $t=-1$ and $s=3$ which then gives $a=\frac{1}{2}$ and $b=2$. But, with these values of $a$ and $b$ you'll notice that the matrix has rank $3$ and so dimension $2$ is not possible.

Can the dimension ever be $3$? Yes, by the above set $a=\frac{1}{2}$ and $b=2$.

Can the dimension ever be $4$? Yes, calculate the determinant and then just find a value for $a$ and $b$ which makes it non-zero, hence the matrix non-singular.

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It comes down to determining how many of the vectors are linearly independent for different choices of $a,b$. For a straightforward case, see what happens when $b=0$, for $v_4$.

Try doing Gaussian elimination on the right-hand side matrix (when $b\neq 0$) and see what happens when $b=0$. Can you find values of $a,b$ so you end up with 1 zero row (yes; use $b=0$)? Can you find values of $a,b$ so you have two zero rows after Gaussian elimination?

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