# find all 2x2 matrices such that a^2=0

so cd = c and d2 = d, obviously d isn't 0 so d is 1 and c is free. Note that in this context A−1 does not mean 1 A. We then get. Use two different nonzero columns for B. I know I can put some variables in B and then multiply AB and then that equation = 0, but I still can't seem to crack it. One way is by doing what you've done and just playing around until you've found something. squaring it. If a2=1, then you know a = 1 (and not -1) from a (ad-bc) = (ad-bc) , above. As you can see, (1) = (3), so this system is dependent, and will have infinitely many solutions. Ok, so, here's some random looks at the equations, without trying to change your basic approach or anything. For a given matrix A, we find all matrices B such that A and B commute, that is, AB=BA. My first thought was to multiply the following matrix by itself: Since I need to get back A, I figure I should then set each element of A2 to the original element I need, and I get this: All I can get from these four equations, however, is something I already know from equations 2 and 3: Any other manipulations I try to do just leads me to the same exact equation. If A^2 = A, it follows that A^2 - A = 0, where 0 is the null or zaro matrix. To do so, we diagonalize the matrix. This chapter is all about transformations, so I still haven't reached invertibility, determinants, etc. I can't think of anything else to do. If you're asking to characterize all of them, it depends on how much theory you have, and with more theory there's better explanations. so from the first paragraph we get ad = bc so if one of a or d is zero one of b or c is also zero. ad-bc = 0 or 1. Factorizing A (A - I) = 0, where I is the identity matrix. If b_1 and b_2 denote the 2x1 columns of a 2x2 matrix B, then the columns of the matrix product AB are precisely the 2x1 vectors Ab_1 and Ab_2. There are many possible solutions. You can probably use an induction proof for this case for integers. Here's the solution worked out if you want it:https://en.wikipedia.org/wiki/Idempotent_matrix#Real_2_.C3.97_2_case, New comments cannot be posted and votes cannot be cast. By doing this, I was able to find a matrix that satisfied A2 = A. I'll check with my professor to make sure it's an acceptable approach tomorrow. I'll take what you said into account as I rework the problem now that I have a matrix satisfying the requirements! Since you want to find a nonzero matrix B with the property that AB is the zero matrix, you want to find two column vectors b_1 and b_2, at least one of which is nonzero, satisfying A b_1 = 0 and A b_2 = 0. An obvious solution to the third equation, then, is. (a-d)(a+d)=(a-d) -- I bet you did this -- but you discarded the a=d case, which you should also analyze. Obviously using four variables is really unwieldy ,so you want to do something to clean it up. You should be able to find 2 of them. These obviously aren't the only solutions in this vein either. Ohhhhh. Find all possible values for the dimension of W. 12. Questions, no matter how basic, will be answered (to the best ability of the online subscribers). For a given matrix A, we find all matrices B such that A and B commute, that is, AB=BA. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. Third: Subtract eqaution 4 from equation 1. Looks like you're using new Reddit on an old browser. But no such real value exists whose square will be negative. b) Find all possible values for the dimension of W. 13. Rank-nullity? Hence, the zero matrix is the only such matrix. This is obviously not a valid deduction to make; a = d = 1, b = c = 0 is a solution, for example, so it is not always true that a+d=1. So let's suppose b = 0. Press question mark to learn the rest of the keyboard shortcuts, https://en.wikipedia.org/wiki/Idempotent_matrix#Real_2_.C3.97_2_case. Are you trying to just find one or to describe all of them? The site may not work properly if you don't, If you do not update your browser, we suggest you visit, Press J to jump to the feed. We have not reached those points, no. If you look at your determinant equation in that light, you get a2={0 or 1}. squaring it. 2×2 determinants can be used to find the area of a parallelogram and to determine invertibility of a 2×2 matrix. A 2×2 determinant is much easier to compute than the determinants of larger matrices, like 3×3 matrices. But that's not a symmetric matrix. The site may not work properly if you don't, If you do not update your browser, we suggest you visit, Press J to jump to the feed. Since there are only 2 idempotent square matrices, you can just try them both for parts a and b. Post all of your math-learning resources here. The next is about invertibility, determinants, image, etc. While there are many matrix calculators online, the simplest one to use that I have come across is this one by Math is Fun. A good way to double check your work if you’re multiplying matrices by hand is to confirm your answers with a matrix calculator. I have a solution, but I got there through a mistake. To solve this problem, we use Gauss-Jordan elimination to solve a system. Let us take the other option: a+d = 0 i.e., a = -d. Now, a^2 = -bc implies c & b are of opposite sign and |c| = a^2/|b|. A given matrix contains a variable. In mathematics, the associative algebra of 2 × 2 real matrices is denoted by M(2, R).Two matrices p and q in M(2, R) have a sum p + q given by matrix addition.The product matrix p q is formed from the dot product of the rows and columns of its factors through matrix multiplication.For = (), let ∗ = (− −). So it's not a valid deduction because I can't divide by a variable unless I'm confident it's non-zero, right? If you left multiply by A-1 you get A = I, so now you know A = I or det(A) = 0. now also we get that this equation is not preserved by scalar multiplication, in fact if something is a member of this set it immediately implies all the scalar multiples of that thing are not in this set. 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