$\dim(\ker(ab)) = \dim(\ker(a)) + \dim(\ker(b))$

Question

Can somebody help me please so solve a self-study problem $V$ is a vector space over $F$ and $a,b$ are in $L(V)$.

Suppose $\ker(a)$ and $\ker(b)$ are finite dimensional and b is surjective.

WTS: $\dim(\ker(ab)) = \dim(\ker(a)) + \dim(\ker(b))$

Answer 1

Note that if $b$ is surjective, and $V$ is finite dimensional, then it also must be an isomorphism (say, by the dimension formula), $\dim\ker b=0$, so the required equation holds because $\dim\ker(ab)=\dim\ker a$.

However, this also holds more generally, with linear maps between (possibly different) vector spaces:
Let $a:V\to W$ and $b:U\to V$ with $b$ surjective, with finite dimensional kernels.
Then, for a $v\in V$, with $b(u)=v$, we have $b^{-1}(v)=u+\ker b$, and if $v_i\in V_i$ are linearly independent, and $u_i$ is their arbitrary preimages, then $u_i$ are linearly independent, too, moreover they are jointly independent from $\ker b$.

Now, take a basis $v_j$ of $\ker a\subseteq V$, and pick a preimage $u_j$ for each $v_j$, and let $u'_k$ be a basis for $\ker b$. Then prove that $u_j, u'_k$ together constitute a basis for $b^{-1}(\ker a)\ =\ \ker(ab)$.

Answer 2

Since

$m = \dim \ker(a) < \infty, \tag 1$

$\ker a$ has a basis

$B_a = \{ a_1, a_2, \ldots, a_m \}; \tag 2$

likewise we have

$n = \dim \ker(b) < \infty, \tag 3$

so $\ker b$ has a basis

$B_b = \{ b_1, b_2, \ldots, b_n \}; \tag 4$

since $b$ is surjective, for each

$a_i \in B_a, 1 \le i \le m, \tag 5$

there is a

$c_i \in V \tag 6$

with

$b(c_i) = a_i \ne 0; \tag 7$

let

$C_b = \{ c_1, c_2, \ldots, c _m\}; \tag 8$

by virtue of (7) and the fact that

$b_i \in B_b \Longrightarrow b(b_i) = 0, \tag 9$

we see that

$B_b \cap C_b = \emptyset; \tag{10}$

I claim that in fact

$\ker(ab) = \langle B_b \rangle \oplus \langle C_b \rangle, \tag{11}$

where $\langle B_b \rangle = \text{span}(B_b)$ etc; for if

$v \in \langle B_b \rangle \oplus \langle C_b \rangle, \tag{12}$

then

$v = \displaystyle \sum_1^n \alpha_i b_i + \sum_1^m \beta_i c_i, \; \alpha_i, \; \beta_i \in F; \tag{13}$

thus

$b(v) = \displaystyle \sum_1^n \alpha_i b(b_i) + \sum_1^m \beta_i b(c_i) = \sum_1^m \beta_i a_i, \tag{14}$

$(ab)(v) = \displaystyle \sum_1^m \beta_i a(a_i) = 0, \tag{15}$

whence

$\langle B_b \rangle \oplus \langle C_b \rangle \subset \ker(ab); \tag{16}$

likewise, for

$v \in \ker(ab) \tag{17}$

we have

$(ab)(v) = 0, \tag{18}$

hence

$bv \in \ker a, \tag{19}$

whence

$bv = \displaystyle \sum_1^m \beta_i a_i = \displaystyle \sum_1^m \beta_i b(c_i) , \tag{20}$

or

$b \left (v -\displaystyle \sum_1^m \beta_i c_i \right ) = 0, \tag{21}$

whence

$v - \displaystyle \sum_1^m \beta_i c_i \in \ker b, \tag{22}$

that is

$v = \displaystyle \sum_1^m \beta_i c_i + \sum_1^m \alpha_i b_i \in \langle B_b \rangle \oplus \langle C_b \rangle ; \tag{23}$

we have thus shown that

$\ker(ab) \subset \langle B_b \rangle \oplus \langle C_b \rangle ; \tag {24}$

combining this with (16) yields (11). It then follows that, since

$\dim \langle C_b \rangle = \dim \langle B_a \rangle = \dim \ker a, \tag{25}$

$\dim \ker(ab) = \dim \ker a + \dim \ker b. \tag{26}$

Nota Bene:

$\dim \langle C_b \rangle = \dim \langle B_a \rangle \tag {27}$

by virtue of the fact that $C_b$ is a linearly independent set, which in turn is a consequence of

$B_a = b(C_b), \tag{28}$

$B_a$ being a basis for $\ker a$. We also observe that (10) implies

$\langle B_b \rangle \cap \langle C_b \rangle =\{0\}, \tag{29}$

since otherwise we would have

$\displaystyle \sum_1^m \alpha_i b_i; \tag{30}$ End of Note.