give answer in 2 step with explanation at the end of each step and final answer at the end:
Let V be a finite dimensional vector space over a field F
,
and φ : V
→
V a linear transformation such

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:
Let V be a finite dimensional vector space over a field F 
,
 ﻿and φ : V 
→
 ﻿V a linear transformation such that φ
^
2
 
=
 ﻿φ
.
 
(
a
)
Show that φ
(
V 
)
 
∩
 ﻿ker φ 
=
 
0
.
 
(
b
)
 ﻿Prove that V 
=
 ﻿φ
(
V
)
 
⊕
 ﻿ker φ
.
 
(
c
)
 ﻿Prove that there is a basis of V such that the matrix of φ with respect to this basis is a diagonal matrix whose entries are all 
0
 ﻿or 
1

VIPSolutions · Accepted Answer

Let's address each part **in two steps** with explanations and then provide the **final answer** at the end of each part.

---

### (a) Show that $ \varphi(V) \cap \ker \varphi = 0 $.

#### **Step 1:** Let $ v \in \varphi(V) \cap \ker \varphi $.

- By definition, $ v \in \varphi(V) $, so there exists $ u \in V $ such that $ v = \varphi(u) $.
- Also, $ v \in \ker \varphi $, so $ \varphi(v) = 0 $.

**Explanation:** This step sets up the facts about any element in the intersection.

---

#### **Step 2:** Show that $ v = 0 $.

- Compute $ \varphi(v) = \varphi(\varphi(u)) = \varphi^2(u) $.
- Since $ \varphi^2 = \varphi $, this means $ \varphi^2(u) = \varphi(u) = v $.
- But we also know $ \varphi(v) = 0 $, so $ v = 0 $.

**Explanation:** Using the property $ \varphi^2 = \varphi $, we see that $ v $ must be zero.

---

**Final Answer (a):**
$$
\varphi(V) \cap \ker \varphi = \{0\}
$$
That is, the intersection contains only the zero vector.

---

### (b) Prove that $ V = \varphi(V) \oplus \ker \varphi $.

#### **Step 1:** Show that every $ v \in V $ can be written as $ v = a + b $ with $ a \in \varphi(V), b \in \ker \varphi $.

- Take any $ v \in V $.
- Let $ a = \varphi(v) \in \varphi(V) $.
- Let $ b = v - \varphi(v) $.
- Check if $ b \in \ker \varphi $: $ \varphi(b) = \varphi(v - \varphi(v)) = \varphi(v) - \varphi^2(v) $.
- Since $ \varphi^2 = \varphi $, $ \varphi^2(v) = \varphi(v) $, so $ \varphi(b) = \varphi(v) - \varphi(v) = 0 $.
- Thus, $ b \in \ker \varphi $.

**Explanation:** Every vector can be decomposed into a part in $ \varphi(V) $ and a part in $ \ker \varphi $.

---

#### **Step 2:** Show that the sum is direct.

- From part (a), $ \varphi(V) \cap \ker \varphi = \{0\} $.
- Thus, the decomposition $ v = a + b $ is unique.

**Explanation:** The intersection being $ \{0\} $ ensures the sum is direct.

---

**Final Answer (b):**
$$
V = \varphi(V) \oplus \ker \varphi
$$
That is, $ V $ is the direct sum of the image and kernel of $ \varphi $.

---

### (c) Prove there is a basis of $ V $ such that the matrix of $ \varphi $ is diagonal with entries 0 or 1.

#### **Step 1:** Choose bases for $ \varphi(V) $ and $ \ker \varphi $.

- Let $ \{e_1, \ldots, e_k\} $ be a basis for $ \varphi(V) $.
- Let $ \{f_1, \ldots, f_m\} $ be a basis for $ \ker \varphi $.
- By (b), $ \{e_1, \ldots, e_k, f_1, \ldots, f_m\} $ is a basis for $ V $.

**Explanation:** Since $ V $ is the direct sum, concatenating bases gives a basis for the whole space.

---

#### **Step 2:** Compute $ \varphi $ on these basis elements.

- For each $ e_i \in \varphi(V) $, there exists $ u_i $ such that $ e_i = \varphi(u_i) $.
- Compute $ \varphi(e_i) = \varphi(\varphi(u_i)) = \varphi^2(u_i) = \varphi(u_i) = e_i $.
- For each $ f_j \in \ker \varphi $, $ \varphi(f_j) = 0 $.
- So, the matrix of $ \varphi $ in this basis is diagonal with $ 1 $'s for the $ e_i $, and $ 0 $'s for the $ f_j $.

**Explanation:** The action of $ \varphi $ on the basis elements gives the required diagonal form.

---

**Final Answer (c):**
There exists a basis of $ V $ such that the matrix of $ \varphi $ is diagonal, with diagonal entries all $ 1 $ (for basis elements of $ \varphi(V) $) and $ 0 $ (for basis elements of $ \ker \varphi $). That is, the matrix looks like
$$
\begin{bmatrix}
I_k & 0 \
0 & 0_m
\end{bmatrix}
$$
where $ I_k $ is the $ k \times k $ identity matrix and $ 0_m $ is the $ m \times m $ zero matrix.

VIPSolutions · Answer

**(a) Final answer:**  
$$
\boxed{
\varphi(V) \cap \ker \varphi = \{0\}
}
$$
The intersection contains only the zero vector.

---

**(b) Final answer:**  
$$
\boxed{
V = \varphi(V) \oplus \ker \varphi
}
$$
The vector space $ V $ is the direct sum of the image and kernel of $ \varphi $.

---

**(c) Final answer:**  
There exists a basis of $ V $ such that the matrix of $ \varphi $ with respect to this basis is diagonal, with entries all $ 0 $ or $ 1 $. Specifically, this basis can be chosen as a union of a basis for $ \varphi(V) $ (on which $ \varphi $ acts as the identity) and a basis for $ \ker \varphi $ (on which $ \varphi $ acts as zero), making the matrix block diagonal with blocks $ I $ and $ 0 $.

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