# Problem in Linear Algebra about dimension of vector space

Let U and V be finite dimensional vector spaces Over $\mathbb R$, Let $L(U,V)$ be the vector space of linear transformations from $U$ to $V$, and Let $W$ be a vector subspace of $U$.

If $Z$= {$T$ $\in L(U,V)$: $T(w)=0$ for all w $\in W$}, then What is the vector space dimension Dim($Z$) of $Z$ in terms of vector space dimension of $U$, $V$, and $W$?

• The definition of $Z$ seems to be incomplete. Perhaps you mean $T(w)=0$ ? – lhf May 6 '16 at 12:51
• well i have made changes. – gaurav May 6 '16 at 13:00
• Hint: write $U=W\oplus \tilde{W}$. – Mathematician 42 May 6 '16 at 13:17

Consider the map $\phi:L(U,V)\rightarrow L(W,V)$ that maps $f\in L(U,V)$ to its restriction to $W$. This map is linear, surjective and $Z$ is exactly its kernel. Hence by the rank-nullity theorem, we get $\dim(Z)=\dim L(U,V)-\dim L(W,V)$.
Now, $\dim L(U,V)=\dim U\cdot\dim V$ and $\dim L(W,V)=\dim W\cdot\dim V$ (see for instance this post: Basis of the space of linear maps between vector spaces). Hence $\dim Z=\dim V(\dim U-\dim W)$.

Hint: Choose a basis of $W$ and extend it to a basis of $U$. Express $T$ in that basis as a matrix. How many degrees of freedom are left?

We claim that $\dim(Z)=[\dim(U)-\dim(W)]\cdot\dim(V)$. Let $\{u_1,u_2,\ldots,u_k\}$ be the basis for $W$, $\{u_1,u_2,\ldots,u_k,u_{k+1},\ldots,u_m\}$ be the extended basis for $U$, and $\{v_1,v_2,\ldots,v_n\}$ be the basis for $V$. Determine the bases for $L(U,V)$ and for $Z$ as below. Given $1\leq i\leq m$ and $1\le j\le n$, there exists exactly one linear map, say $T_{ij}\in L(U,V)$, satisfying (see the Lemma shown below) $$T_{ij}(u_r)=\delta_{ir}v_j,\quad\forall 1\le r\le m,$$ where $\delta_{ir}$ is the Kronecker delta function. It is easy to see that (leave it to check) $\beta=\{T_{ij}:1\le i\le m\mbox{ and }1\le j\le n\}$ is the basis for $L(U,V)$. Now, given $T\in Z$, we first assume $T=\displaystyle\sum_{i,\,j}a_{ij}T_{ij}$ for some scalars $a_{ij}$. Since $u_1,u_2,\ldots,u_k\in W$, we have $${\it 0}=T(u_r)=\sum_{i,\,j}a_{ij}T_{ij}(u_r)=\sum_{i,\,j}a_{ij}\delta_{ir}v_j=\sum_{j=1}^na_{rj}v_j\quad \mbox{for }1\le r\le k.$$ Also, since $\{v_1,v_2,\ldots,v_n\}$ is linearly independent, $a_{r1}=a_{r2}=\cdots=a_{rn}=0$ for all $1\le r\le k$. Therefore $$T=\sum_{i,\, j}a_{ij}T_{ij}=\sum_{i=k+1}^m\sum_{j=1}^na_{ij}T_{ij},$$ and thus $\gamma=\{T_{ij}:k+1\le i\le m\mbox{ and }1\le j\le n\}$ generates $Z$. Finally, since $\gamma$ is clearly linearly independent, we conclude that $\gamma$ is a basis for $Z$. Hence $$\dim(Z)=(m-k)\cdot n=[\dim(U)-\dim(W)]\cdot\dim(V).$$

Lemma. Let $U$ and $V$ be vector spaces, and suppose that $\{u_1,u_2,\ldots,u_m\}$ is a basis for $U$. For $v_1,v_2,\ldots,v_m\in V$, there exists exactly one linear map $T\in L(U,V)$ such that $T(u_i)=v_i$ for $i=1,2,\ldots,m$.

Proof. Let $x\in U$, then $x=\displaystyle\sum_{i=1}^ma_iu_i$, for some unique scalars $a_i$. Define $T:U\rightarrow V$ by $$T(x)=\sum_{i=1}^ma_iv_i.$$ Then it is clear that $T(u_i)=v_i$ for each $i$. To check $T$ is linear, given $u,u'\in U$ and a scalar $c$, then we may write $$u=\sum_{i=1}^mb_iu_i\quad\mbox{and}\quad u'=\sum_{i=1}^mb_i'u_i$$ for some scalars $b_i$ and $b_i'$. Thus $cu+u'=\displaystyle\sum_{i=1}^m(cb_i+b_i')u_i$ and so $$T(cu+u')=\sum_{i=1}^m(cb_i+b_i')v_i =c\sum_{i=1}^mb_iv_i+\sum_{i=1}^mb'_iv_i=c\,T(u)+T(u').$$ Hence $T$ is linear. Next, to check $T$ is unique, suppose that there is another $T'\in L(U,V)$ such that $T'(u_i)=v_i$ for each $i$. Then for $x\in U$ with $x=\displaystyle\sum_{i=1}^ma_iu_i$, we have $$T'(x)=\sum_{i=1}^ma_iT'(u_i)=\sum_{i=1}^ma_iv_i =\sum_{i=1}^ma_iT(u_i)=T(x).$$ Hence $T'=T$.

• thanks..got it but proof is lenghty – gaurav May 7 '16 at 3:01
• Though it is lengthy, I've tried to make it more easy to understand. Before the gray line our principal proof, and after that is an additional appendix proving one non-trivial property which is used in this problem. – Solumilkyu May 7 '16 at 7:14