# When does (a+b)^2 = a^2 + b^2 hold true?

#### abcd1234

##### New member
I am trying to find out as many different possibilities where this holds true including for numbers, matrices, vectors and anything else.

The only case I have got this will hold true if 2ab=0 - i.e. when a=0 or b=0 or if both a and b are 0
I think this covers the numbers case?
I have no idea where to start on matrices, vectors etc.

Any help would be much appreciated.

#### Jomo

##### Elite Member
I am trying to find out as many different possibilities where this holds true including for numbers, matrices, vectors and anything else.

The only case I have got this will hold true if 2ab=0 - i.e. when a=0 or b=0 or if both a and b are 0
I think this covers the numbers case?
I have no idea where to start on matrices, vectors etc.

Any help would be much appreciated.
real numbers:If (a+b)2 = a2+b2 + 2ab = a2+b2, then 2ab=0 so ab=0 so a=0 or b=0 ( just as you said)
Matrices: If (A+B)2 = (A+B)(A+B) = A2 + AB + BA + B2 = A2 + B2, then AB + BA = 0. Try to figure out the conditions on A and B.
Vectors: Let V = < v1, v2, ... ,vn> and W = < w1, w2, ... ,wn>. Compute (V + W)2 and find the conditions on V and W such that it equals V2 + W2

#### abcd1234

##### New member
real numbers:If (a+b)2 = a2+b2 + 2ab = a2+b2, then 2ab=0 so ab=0 so a=0 or b=0 ( just as you said)
Matrices: If (A+B)2 = (A+B)(A+B) = A2 + AB + BA + B2 = A2 + B2, then AB + BA = 0. Try to figure out the conditions on A and B.
Vectors: Let V = < v1, v2, ... ,vn> and W = < w1, w2, ... ,wn>. Compute (V + W)2 and find the conditions on V and W such that it equals V2 + W2
Thank you so much Jomo.
I think that hint for the matrices has really helped get me started:

Matrices
Working backwards - the matrices AB and BA can only be added together if they are of the same size (i.e. have the same dimensions).
Now the product matrices AB and BA are always defined when A is mxn AND B is nxm.
BUT the matrices AB and BA will be of different dimensions and not the same size (so contradicting the first line) - i.e. in this case we cannot AB and BA
So A and B must each be matrices of the same dimensions (i.e. m = n)
So A and B are square matrices (in order for AB and BA to be defined).
From there, I believe the only square matrices for A and B that can be satisfied are if either A is the 0 matrix, or if B is the 0 matrix, or if both A and B are the 0 matrix, all with the condition that A and B are of the same dimensions.

Please let me know if this is complete/or have I made a huge error (as I'm usually not great at solving things myself!)

Vectors
Again, two vectors of different sizes cannot be added and hence the vectors A and B must be of the same dimensions.
Using the vectors a = < a1, a2, ... , an> and b = < b1, b2, ... ,bn>

I found that (a+b)2 = < a12+2a1b1+b22 , a22+2a2b2+b22​ ,..., an2+2anbn+bn2​ >
If (a+b)2 = a2 +b2, this vector must equal < a12+b22 , a22+b22​ ,..., an2+bn2​ >
So 2a1b1=0, 2a2b2=0,.....2anbn​=0
So either ak=0, or bk​=0, or both ak and bk​=0 (for any integer 1<=k<=n)

Again is this correct?

Finally, is there anything else I could look at other than matrices, vectors and real numbers?

Thank you so much for all your help!!

#### ksdhart2

##### Full Member
Matrices
Working backwards - the matrices AB and BA can only be added together if they are of the same size (i.e. have the same dimensions).
Now the product matrices AB and BA are always defined when A is mxn AND B is nxm.
BUT the matrices AB and BA will be of different dimensions and not the same size (so contradicting the first line) - i.e. in this case we cannot AB and BA
So A and B must each be matrices of the same dimensions (i.e. m = n)
So A and B are square matrices (in order for AB and BA to be defined).
From there, I believe the only square matrices for A and B that can be satisfied are if either A is the 0 matrix, or if B is the 0 matrix, or if both A and B are the 0 matrix, all with the condition that A and B are of the same dimensions.

Please let me know if this is complete/or have I made a huge error (as I'm usually not great at solving things myself!)
Your conclusion that A and B must be square matrices of the same size is correct, although perhaps an easier way to get there would be to consider the problem itself. We're asked to find any cases where $$\displaystyle (A + B)^2 = A^2 + B^2$$. Well, in order for this equality to hold true, both sides of it must be defined. Squaring a matrix is the same as multiplying by itself, ad the rules of multiplication mean that only square matrices can be multiplied by itself. These two resulting matrices must then be added together, which can only be done if they have the same dimensions. Hence, A and B must be same-sized square matrices.

However, it's not true that one (or both) of A and B must be the zero matrix. Certainly, if one (or both) are the zero matrix, the required equality $$\displaystyle AB + BA = 0$$ holds, but those aren't the only conditions under which it holds. Consider this example:

$$\displaystyle \begin{pmatrix}0&0\\ 3&4\end{pmatrix}\begin{pmatrix}1&0\\ -\frac{3}{4}&0\end{pmatrix}+\begin{pmatrix}1&0\\ -\frac{3}{4}&0\end{pmatrix}\begin{pmatrix}0&0\\ 3&4\end{pmatrix}=\begin{pmatrix}0&0\\ 0&0\end{pmatrix}$$

Finally, is there anything else I could look at other than matrices, vectors and real numbers?
The stipulation of "and anything else." is really quite broad. You might try investigating complex numbers. Let $$\displaystyle A = a + bi$$ and $$\displaystyle B = c + di$$ and see where that leads you. A lot depends on your level of familiarity with mathematics, as to what might be appropriate to explore. There are more "exotic" things you can play around with, such as quaternions, octonions, etc. But those can quickly get out of hand, as they start to lose properties such as commutatitvity, associativity, and introduce zero divisors (i.e. two non-zero elements A and B can multiply together to equal 0)

#### Jomo

##### Elite Member
However, it's not true that one (or both) of A and B must be the zero matrix. Certainly, if one (or both) are the zero matrix, the required equality $$\displaystyle AB + BA = 0$$ holds, but those aren't the only conditions under which it holds. Consider this example:

$$\displaystyle \begin{pmatrix}0&0\\ 3&4\end{pmatrix}\begin{pmatrix}1&0\\ -\frac{3}{4}&0\end{pmatrix}+\begin{pmatrix}1&0\\ -\frac{3}{4}&0\end{pmatrix}\begin{pmatrix}0&0\\ 3&4\end{pmatrix}=\begin{pmatrix}0&0\\ 0&0\end{pmatrix}$$
Your example clearly showed that AB + BA = 0 with A or B not being 0. However your AB and BA both were 0. Is that a necessary condition for AB + BA = 0? That an honest question from me because I am not sure yet if that is true. Can you please show a proof if it is a necessary condition. Thanks.

#### Subhotosh Khan

##### Super Moderator
Staff member
Your example clearly showed that AB + BA = 0 with A or B not being 0. However your AB and BA both were 0. Is that a necessary condition for AB + BA = 0? That an honest question from me because I am not sure yet if that is true. Can you please show a proof if it is a necessary condition. Thanks.
In that example:

|A| = 0 = |B|

#### Jomo

##### Elite Member
In that example:

|A| = 0 = |B|
Yes, I noticed that. So now I'll ask for AB + BA = 0, must it be that |A| = 0 = |B| and not AB = 0 = BA?