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I'm currently reading book on linear algebra where author presents concept of the "mathematical induction". The author explains 3 steps of mathematical induction:

Step 1. Check the result is true for some base case such as $n = 1$

Step 2. Assume the result is true for $n = k$.

Step 3. Prove the result for $n = k + 1$ by using steps $1$ and $2$

Honestly, I'm confused about step 2. Namely, how can we assume that the result is true for some arbitrary value $k$? Isn't it what we are actually required to prove?

In an attempt to make my question clearer, I'll provide an example:

The author presents following proposition:

Proposition 1. Let $\mathbf{A}_{1}\mathbf{A}_{2}...\mathbf{A}_{n-1}\mathbf{A}_{n}$ be matrices of the approriate size so that they can be multiplied. Then we have $(\mathbf{A}_{1}\mathbf{A}_{2}...\mathbf{A}_{n-1}\mathbf{A}_{n})^{T} = \mathbf{A}_{n}^{T}\mathbf{A}_{n-1}^{T}...\mathbf{A}_{2}^{T}\mathbf{A}_{1}^{T}$

His proof:

Check the result for $n = 2$: $$\tag1(\mathbf A_{1}\mathbf A_{2})^{T} = \mathbf A_{2}^{T}\mathbf A_{1}^{T}$$

Assume result is true for $n = k$:

$$\tag2(\mathbf{A}_{1}\mathbf{A}_{2}...\mathbf{A}_{k-1}\mathbf{A}_{k})^{T} = \mathbf{A}_{k}^{T}\mathbf{A}_{k-1}^{T}...\mathbf{A}_{2}^{T}\mathbf{A}_{1}^{T}$$

Required to prove the result for $n = k + 1$, that is we need to prove:

$$\tag3(\mathbf{A}_{1}\mathbf{A}_{2}...\mathbf{A}_{k-1}\mathbf{A}_{k}\mathbf{A}_{k+1})^{T} = \mathbf{A}_{k+1}^{T}\mathbf{A}_{k}^{T}\mathbf{A}_{k-1}^{T}...\mathbf{A}_{2}^{T}\mathbf{A}_{1}^{T}$$

Expanding the Left Hand Side $(\mathbf{A}_{1}\mathbf{A}_{2}...\mathbf{A}_{k-1}\mathbf{A}_{k}\mathbf{A}_{k+1})^{T}$ gives:

$$(\mathbf{A}_{1}\mathbf{A}_{2}...\mathbf{A}_{k-1}\mathbf{A}_{k}\mathbf{A}_{k+1})^{T} = ((\mathbf{A}_{1}\mathbf{A}_{2}...\mathbf{A}_{k-1}\mathbf{A}_{k})\mathbf{A}_{k+1})^{T} = \mathbf{A}_{k+1}^{T}(\mathbf{A}_{1}\mathbf{A}_{2}...\mathbf{A}_{k-1}\mathbf{A}_{k})^{T}$$

Using the result we obtained in $(2)$: $$\mathbf{A}_{k+1}^{T}(\mathbf{A}_{1}\mathbf{A}_{2}...\mathbf{A}_{k-1}\mathbf{A}_{k})^{T} = \mathbf{A}_{k+1}^{T}\mathbf{A}_{k}^{T}\mathbf{A}_{k-1}^{T}...\mathbf{A}_{2}^{T}\mathbf{A}_{1}^{T}$$

As desired. $\Box$

OK, so now I repeat my question:

How can we assume that proposition holds true for some arbitrary value $k$, if this is what we actually are required to prove? We didn't actually obtain the result in (2), we just assumed that it is true.

I'd thought that the proof would need to look like this:

  1. Check the result for $n = 2$: $$\tag1(\mathbf A_{1}\mathbf A_{2})^{T} = \mathbf A_{2}^{T}\mathbf A_{1}^{T}$$

  2. Now pick some arbitrary value $n = k$. Required to prove that proposition holds for $k$, i.e:

$$\tag2(\mathbf{A}_{1}\mathbf{A}_{2}...\mathbf{A}_{k-1}\mathbf{A}_{k})^{T} = \mathbf{A}_{k}^{T}\mathbf{A}_{k-1}^{T}...\mathbf{A}_{2}^{T}\mathbf{A}_{1}^{T}$$

Using property we discovered in (1) $$\tag2(\mathbf{A}_{1}\mathbf{A}_{2}...\mathbf{A}_{k-1}\mathbf{A}_{k})^{T} = ((\mathbf{A}_{1}\mathbf{A}_{2}...\mathbf{A}_{k-1})\mathbf{A}_{k})^{T} = \mathbf{A}_{k}^{T} (\mathbf{A}_{1}\mathbf{A}_{2}...\mathbf{A}_{k-1})^{T}$$

Repeat the same step one more time

$$\mathbf{A}_{k}^{T} ((\mathbf{A}_{1}\mathbf{A}_{2}...)\mathbf{A}_{k-1})^{T} = \mathbf{A}_{k}^{T} \mathbf{A}_{k-1}^{T}(\mathbf{A}_{1}\mathbf{A}_{2}...)^{T}$$

Using the same process for the remaining matrices within brackets, finally we arrive at:

$$\mathbf{A}_{k}^{T}\mathbf{A}_{k-1}^{T}...\mathbf{A}_{2}^{T}\mathbf{A}_{1}^{T}$$

As desired. $\Box$

What am I missing?

Community
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Stokolos Ilya
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  • Have you proved anything by induction earlier? To prove that $P(n)$ is true for all $n \geq 1$ you prove it for $n=1$ , assume that $P(k)$ is true and prove that P((K+1)$ is true. – Kavi Rama Murthy Sep 03 '19 at 08:03
  • Usually we have ony two steps : (i) Base case : prove that the sought property holds for $n=1$, and (ii) Induction step : assume that the property holds for $n=k$ and prove that it holds for $n=k+1$. This is the correct way to use te general principle (or axiom) of Mathematical Induction. – Mauro ALLEGRANZA Sep 03 '19 at 08:04
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    See the post The concept of Mathematical Induction as well as How can Mathematical Induction prove something – Mauro ALLEGRANZA Sep 03 '19 at 08:16
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    Good question, we are trying to prove for arbitrary value k. Look at is this way: Let's assume that it's true for n = k. Then, if we can prove using this assumption that it's true for n = k+1, then we can then say k = 1, which means n = k + 1 = 2 is true, so n = 2 + 1 = 3 is true, so that n = 3 + 1 = 4 is true... – battletwink69 Sep 03 '19 at 08:17
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    You have not assumed that the statement is true for all $n$. You are trying to show that if the statement is true for $k$, then it will also be true for $k+1$. If this is proven, since you have proven that the statement is true for the base case $k=2$, it will be true for $k+1=3$; since it is true for $3$ it will be true for $4$ and so on. – Shubham Johri Sep 03 '19 at 08:31

2 Answers2

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This is the structure of all proofs by induction.

Show that the proposition is true in some base case. Assume that it is true in some general case (the inductive hypothesis). Based on the hypothesis, show that when it holds for some value of n, that it also holds for the value $n+1$

i.e. you have shown that the proposition holds when $k = 2$

And whenever it holds for some $k,$ it must also hold for $k+1.$ So, it must hold when k= 3, since it holds when $k = 2$

And it must hold when $k=4$ since it holds when $k = 3$

And following this logic along, it holds for all integers greater than or equal to 2.

https://en.wikipedia.org/wiki/Mathematical_induction

Doug M
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Principle of Mathematical Induction is just the mathematical representation for the game of Dominoes. How do you ensure that all the dominoes fall? You need two conditions:

1) The first one falls.

2) Whenever one falls, it pushes the next one.

The basis step represents the first one and the hypothesis and inductive steps together represent the second step. What does they say? They say that when the proposition is true for a natural number, it is true for the next natural number.

This ensures that the given proposition is true for all natural numbers. Otherwise, the first tile that couldn't fall was not pushed by the previous one.

Hope it helps:)

Martund
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