Questions in Proposition $7.23$ of Textbook beginning in Algebraic Geometry by Clader and Ross

I was self studying Algebraic Geometry from the textbook of Algebraic Geometry by Clader and Ross: Beginning in Algebraic Geometry. I have questions on page $206-207$ of the textbook.

Background information: If $f \in K[x_1,...,x_n]$ and $a=(a_1,...,a_n)\in \mathbb{A}^n$ , then the linearization of $f$ at $a$ is defined by $L_a(f) = f(a) +\sum_{i=1}^n (\frac{\partial f}{\partial{x_i}}(a)(x_i-a_i)) \in K[x_1,...,x_n]$.

Let $X\in \mathbb{A}^n$ be an affine variety and $a\in X$ . The linearlization of $X$ at $a$ is $L_aX= V(${$L_af| f\in I(X)$}$) \subseteq \mathbb{A}^n$.

Let $X \subseteq \mathbb{A}^n$ be an affine variety and $a=(a_1,...,a_n)\in X$ . For any $b=(b_1,....,b_n)\in L_aX$, the tangent vector associated to $b$ is defined by ${ab}^{\rightarrow}=(b_1-a_1,...,b_n-a_n) \in K^n$. The tangent space of $X$ at $a$ is the collection of tangent vectors: $T_aX = ${${ab}^{\rightarrow}|b \in L_aX$}$\subseteq K^n$.

Proposition $7.23$: Let $X \subset \mathbb{A}^n$ be an irreducible affine variety and $a\in X$. Then $dim(T_a X)\geq dimX$

I completely understand the proof of this subsequent Lemma:$7.24$ If $X \subset \mathbb{A}^n$ is an irreducible affine variety and $a\in X$ , then dim$(X)=0$ iff dim$(T_aX)=0$.

Proof of Proposition $7.23$ We prove the proposition by using induction on $dim(X)$.

Base Case If dimX=0. This case is clear to me.

Induction Step Let $X\subseteq \mathbb{A}^n$ be an irreducible affine variety of positive dimension and suppose that the inequality of the proposition holds for all irreducible affine varieties of dimension dim(X)$-1$ . Let $a=(a_1,...,a_n)\in X$ .

Since we have assumed that $dim(X)>0$ , the if direction of Lemma $7.24$ implies that $T_a(X) \supsetneq${$0$} which is equivalent to $L_aX\supsetneq ${$a$}. I am unable to prove this equivalence, please help me Thus we can choose a point $b \in L_aX ${$a$}. The two points $a,b \in \mathbb{A}^n$ must differ in atleast one co-ordinate. WLOG, assume that they differ in $1$st co-ordinate and define the hyperplane $H=V(x_1-a_1)\subseteq \mathbb{A}^n$.

Set $Y= X\cap H$. Notice first that $Y$ can't be all of $X$. Indeed if $Y=X$, then $X\subset H$ , which would imply that $L_aX\subseteq H$. But $H$ was chosen specifically so that $b\in L_aX$ but $b\notin H$, so $L_aX\nsubseteq H$. Thus $Y\neq X$ and since $a\in Y$ we conclude that $\emptyset\nsubseteq Y \nsubseteq X,$, from which the strong form of the fundamental theorem of dimension theory implies that every irreducible component of $Y$ has dimension $dim($X$)-1$.

Let $Z$ be an irreducible component of $Y$that contains $a$. Since $Z \subseteq X$ and $Z\subseteq H$, we have $L_aZ\subseteq L_aX$ and $L_aZ \subseteq H$.

I am not able to deduce the following line $L_aZ \subseteq L_AX \cap H \subsetneq L_a X,$ and

from this how does it follows that $T_aZ\subsetneq T_aX$, so $dim(T_aZ )< dim(T_aX).$

Rest of the proof is clear to me.

Please help me with these question.

I shall be extremely grateful.