Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 1 / 9F (Linear Algebra)

False. A left inverse would force $\alpha$ to be injective: if $\alpha u=0$, then $u=\beta \alpha u=0$. Take the surjective linear map $\alpha :{\mathbb{R}}^2\to \mathbb{R}$ given by $\alpha (x,y)=x$. Since $(0,1)$ belongs to its kernel, every composite $\beta \alpha$ annihilates $(0,1)$ and cannot be the identity map on ${\mathbb{R}}^2$.

True. Choose a basis $v_1,\dots ,v_r$ of $V$. By surjective function, choose $u_i\in U$ with $\alpha u_i=v_i$. Define the linear map $\beta$ by $\beta v_i=u_i$ and extend by linearity. Then $\alpha \beta v_i=v_i$ on every basis vector, and thereforeThis constructs a right inverse and does not require $\alpha$ to be injective. If $V$ is the zero vector space, the unique zero map is the required right inverse.

True. Use the lifted basis $u_1,\dots ,u_r$ above and put $W=\mathrm{span}\{u_1,\dots ,u_r\}$. If $\sum c_i{u}_i=0$, applying $\alpha$ gives $\sum c_i{v}_i=0$, so all $c_i=0$. Thus the $u_i$ form a basis of $W$, and $\alpha {|}_W$ takes this basis to the basis of $V$. It is a linear isomorphism.

In fact the right inverse gives the stronger splittingFor every $u\in U$, write $u=(u-\beta \alpha u)+\beta \alpha u$; the first term is in the kernel and the second in $W$. Their intersection is zero because $\alpha {|}_W$ is injective. This is a direct sum decomposition.

False. Take $U={\mathbb{R}}^2$, $V=\mathbb{R}$, $\alpha (x,y)=x+y$, $X=\mathrm{span}\{(1,0)\}$, and $Y=\mathrm{span}\{(0,1)\}$. Then $U=X\oplus Y$ is a direct sum, but $\alpha (X)=\alpha (Y)=\mathbb{R}$, so their intersection is not zero. Their ordinary sum is $V$, as always follows from surjective function and $U=X+Y$; the failure is precisely the direct sum assertion.

False. Take $\alpha (x,y)=x$, $X=\mathrm{span}\{(1,0)\}$, and $Y=\{0\}$. Then $V=\alpha (X)\oplus \alpha (Y)=\mathbb{R}\oplus \{0\}$, but $X+Y\ne U$ because $(0,1)$ is missing. Thus the premise does not force spanning. It need not force zero intersection either: with the same $\alpha$, take $X=U$ and $Y=\ker \alpha$. Their images still have a direct sum, while $X\cap Y=Y\ne \{0\}$.