Vector Spaces

$\text{Find basis for the kernel and range of:}$ $T(x,y,z=(2x+3y+5z,x+2y+3z,-3x+y+z)$

The kernel you want to find every vector who’s output is the zero vector

Kernel of T:

• $2x+3y+5z=0$
• $x+2y+3z=0$
• $-3x+y+z=0$

this is the same thing as this:

$\left[\begin{array}{ccc}2& 3& 5\\ 1& 2& 3\\ -3& 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

Lets get RREF, $\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right]$

Lemma ${\text{Kernel}}_T={\text{Nullspace}}_A$

Solution in vector form: $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-z\\ -z\\ -z\end{array}\right]=z\left[\begin{array}{c}-1\\ -1\\ -1\end{array}\right]$

Thus KernelT = Nullspace A = Span $\{[-1,-1,-1]\}$ which is a line in ${\mathbb{R}}^3$ ⇐ Basis of kernel Range T = Col A = Span $\{\left[\begin{array}{c}2\\ -1\\ 3\end{array}\right],\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right],\left[\begin{array}{c}5\\ 3\\ 1\end{array}\right]\}$

The last one is redundant so this is actually the span of the first two, making this a plane in ${\mathbb{R}}^3$

the first two are the basis of range T

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Things to check on a subspace

1. $\vec{O}$ is in the set
2. $\overrightarrow{w_1},\overrightarrow{w_2}$ in the set implies $\overrightarrow{w_1}+\overrightarrow{w_2}$ is in the set
3. If $c$ is.a real number and $\vec{w}$ is in the set, then $c\vec{w}$ is must be in the set

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New: The dimension of a vector space

• The dimension of a vector space V is equal to the number of vectors in any basis

If $W=\text{Span}\{v,v_2,\dots v_k\}$ then W is a basis/subspace

Null space of A is all solutions = zero

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Dimension in ${\mathbb{R}}^n$ standard basis is $\{\left[\begin{array}{c}1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\end{array}\right]\}=\{{\vec{e}}_1,{\vec{e}}_2\}$ Thm: if $W\subseteq V$ and W, V are vector spaces (W in V) then $\text{dim}W\le \text{dim}V$

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2/13/25 2/18/25