# Linear Algebra - Find $\lambda$ for linear transformation

$U=\operatorname{Sp}\,\{(1,1,1,1),(\lambda,2\lambda+1,1,2\lambda+1),(\lambda,\lambda,1,\lambda^2)\}$

Find all values for $\lambda$ so $\dim U=3$

So there is a linear transformation from $\mathbb{R}^4 \rightarrow \mathbb{R}^4$ that applies $\ker T = U$ and $T(1,1,2,4)=(1,0,0,0).$

My solution (Not full)

$\dim U=3 \rightarrow$ Only for $\lambda \neq -1$ or $\lambda \neq 1$.

for $\lambda=0 \rightarrow \ker T=\operatorname{Sp}\,\{(1,1,1,1),(0,1,1,1),(0,0,1,0)\} \rightarrow$

$\ker T=\operatorname{Sp}\{(1,0,0,0),(0,1,0,1),(0,0,1,0)\}$

Therefore $\dim \operatorname{Im}\,T=1$

$\operatorname{Im}\,T=\{(0,1,0,0)\}$ or $\{(0,0,0,1)\}$

Any ideas how to continue from here?

• matrixcalc.org if you want to solve any equation of this Nov 16, 2021 at 5:54

Let your vectors be $v_1,v_2,v_3$, and assume $\dim U<3$. Since $v_1$ and $v_2$ are linearly independent, we have $v_3=av_1+bv_2$. The first two coordinates in $v_3$ and $v_1$ (but not in $v_2$ unless $\lambda=-1$) coincide. We have $\lambda=-1$ or $b=0$; in the latter case we have $a=1$ and $\lambda=1$. Finally, note that, for $\lambda=\pm1$, the dimension of $U$ is actually less than $3$.