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

Considering we have to multiply entry 1-2 with entry 2-1, this would mean we're mulitplying the same value if the matrix is symmetric, i.e. A square root of a 2×2 matrix M is another 2×2 matrix R such that M = R 2, where R 2 stands for the matrix product of R with itself. Based on where we are in the book, this is the approach I'm going to use. Assuming you don't need to find every such matrix, just one, a trick you can do is try setting some of these variables equal to 0. 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. 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. 2×2 determinants can be used to find the area of a parallelogram and to determine invertibility of a 2×2 matrix. Note that in this context A−1 does not mean 1 A. Always double-check in the end anyway. More from my site. Factorizing A (A - I) = 0, where I is the identity matrix. 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 ∗ = (− −). If A^2 = A, it follows that A^2 - A = 0, where 0 is the null or zaro matrix. For a given 2 by 2 matrix, we find all the square root matrices. Hi, If you look at your definition of idempotent A^2=A, then you can actually solve this for A and find *all* idempotent square matrices. You get a2 - d2 = a - d, e.g. We then get, We then get that a and d are both plus or minus 1. Construct a 2x2 matrix B such that AB is the zero matrix. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … In addition to multiplying a matrix by a scalar, we can multiply two matrices. So it's not a valid deduction because I can't divide by a variable unless I'm confident it's non-zero, right? Similarly, if you add equations 2 and 3 together, you get (a+d)(b+c)=(b+c), which, once again, gives you an alternative b+c=0 case (EDIT: This is redundant with the b=c=0 cases, though, so kinda not useful). It might be that there are no such matrices, of course, but then you'll come up with a contradiction and you'll know there was a mistake. Matrix Calculator. Note: If the entries are allowed to be complex numbers, then there are more solutions. A matrix, in a mathematical context, is a rectangular array of numbers, symbols, or expressions that are arranged in rows and columns. Let A denote the given matrix. Are you trying to just find one or to describe all of them? Well my first inclination was to find them all so I could become experienced solving these sorts of questions, but I'm not faring too well. Post all of your math-learning resources here. All symmetric 2x2 matrices will be of the form. Find nonzero matrices A;B;Csuch that AC= BCand A6= B. You should analyze those cases as well. New comments cannot be posted and votes cannot be cast, More posts from the cheatatmathhomework community, Continue browsing in r/cheatatmathhomework, Looks like you're using new Reddit on an old browser. A given matrix contains a variable. To do so, we diagonalize the matrix. Let's transform this homogeneous system into a matrix. But no such real value exists whose square will be negative. I have a solution, but I got there through a mistake. For a given matrix A, we find all matrices B such that A and B commute, that is, AB=BA. Theinverseofa2× 2 matrix The inverseof a 2× 2 matrix A, is another 2× 2 matrix denoted by A−1 with the property that AA−1 = A−1A = I where I is the 2× 2 identity matrix 1 0 0 1!. So let's suppose b = 0. Hence, the zero matrix is the only such matrix. a could have been - 1 and d 1, or we could have set a different variable equal to 0, or we could have set it equal to something else and seen what happened. This chapter is all about transformations. A matrix is said to be skew-symmetric if AT = A. Problems and Solutions of Linear Algebra in Mathematics. Matrix A matrix is usually characterised by the number of columns and number of rows it has. For the case where det(A) = 1, you can easily show that there is only one instance of this. That is, multiplying a matrix by its inverse produces an identity matrix. Rank-nullity? If a=0, then you have the zero matrix. Can you find 2x2 matrices A and B such that AB is the zero matrix, but neither A nor B are the zero matrix? So let's suppose b = 0. The square of a matrix A, is given by {eq}A^2 = A*A {/eq} For any n X n matrix to have its square as a zero matrix, it's first necessary that it has all its main diagonal entries as zeroes. This chapter is all about transformations, so I still haven't reached invertibility, determinants, etc. For instance, if we assume b is 0 we can use the equations to find one example of an idempotent matrix with 0 in the top right. That's the question. Which means you're done with that case - you have two degrees of freedom, a and either b or c. If you backtrack, and say a+d is not 1, then you get a=d (see "Third" above), b=0, and c=0. 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. 10 True of False Problems about Nonsingular / Invertible Matrices 10 questions about nonsingular matrices, invertible matrices, and linearly independent vectors. squaring it. Squaring this matrix and setting it to 0, we end up with three equations: a2 +b2 =0 ab+bc=0 b2 +c2 =0 Which values of a,b,c satisfy these equations? 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. So I have this homework problem that I thought would be rather simple and straightforward, but I was wrong. The 1-2 and 2-1 entries have to be the same in a 2x2 symmetric matrix. ad-bc = 0 or 1. Solution: Many solutions, but one example is A= 1 1 1 1 1. So if entry 1-1 is a, the first multiplication is a*a = a2. To find a 2×2 determinant we use a simple formula that uses the entries of the 2×2 matrix. Press question mark to learn the rest of the keyboard shortcuts. How about image space? There are many possible solutions. 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. If you look at your determinant equation in that light, you get a2={0 or 1}. So am I right in saying that no symmetric matrix A exists (other than the zero matrix, of course) such that A2 = 0? The quiz is designed to test your understanding of the basic properties of these topics. What about for a 3x3 matrix, or an arbitrary nxn matrix? The first thing I'll point out is that if you only have to find one such example, the identity matrix clearly works. Questions, no matter how basic, will be answered (to the best ability of the online subscribers). The examples above illustrated how to multiply 2×2 matrices by hand. Now, some more clever ideas: take equation 1 times d minus equation 2 times c. This makes the resulting right side equal to the determinant: Which turns into either the determinant is 0 (which we already had as a possibility) or a = 1, which is exciting. You're very clever! Starting with a^2 + b^2 = 0, b(a+c) = 0, and b^2 + c^2 = 0: Since det(A)2 =det(A), that should give you det(A) = 0 or 1, e.g. Have you gotten to vector spaces and basis? We don't need an augmented matrix… The next value to be added to this must add up to -a2. Thank you so much for your input. An obvious solution to the third equation, then, is. For instance, if we assume b is 0 we can use the equations to find one example of an idempotent matrix with 0 in the top right. That is, if I perform the division to get a + d = 1 from equations 2 and 3, it must mean that both b and c are non-zero, but then we find ourselves with the equation bc = 0, thereby indicating our division was flawed. As you can see, (1) = (3), so this system is dependent, and will have infinitely many solutions. A good way to double check your work if you’re multiplying matrices by hand is to confirm your answers with a matrix calculator. Help please. If you plug that into equations 1 and 4, you should see that they are satisfied; equations 2 and 3 were already satisfied with a + d = 1. Ok, so, here's some random looks at the equations, without trying to change your basic approach or anything. 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. Let U and V both be two-dimensional subspaces of R5, and deﬁne the set W := U+V as the set of all vectors w = u+v where u ∈ U and v ∈ V can be any vectors. Is this the right conclusion to make or am I still missing it? Press question mark to learn the rest of the keyboard shortcuts, https://en.wikipedia.org/wiki/Idempotent_matrix#Real_2_.C3.97_2_case. so cd = c and d2 = d, obviously d isn't 0 so d is 1 and c is free. I don't think there is one other than the zero matrix itself. Find all possible values for the dimension of W. 12. 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|. So if entry 1-1 is a, the first multiplication is a*a = a 2. We want to determine the all values of it so that the matrix nonsingular. 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. If you assume a+d = 1, and plug that into (EDIT(clarity): your equation 4), you get bc + (1-a)2 = 1-a, which simplifies to bc + a(a-1) = 0; since a-1=-d, this is actually bc-ad=0 for that case, which means that if a+d = 1, the determinant must be 0, e.g. (a-d)(a+d)=(a-d) -- I bet you did this -- but you discarded the a=d case, which you should also analyze. Then q q * = q * q = (ad − bc) I, where I is the 2 × 2 identity matrix. The next value to be added to this must add up to -a 2. squaring it. Looks like you're using new Reddit on an old browser. We use the elementary row operations to solve it. Eigenvalues and characteristic polynomials? We have not reached those points, no. First, a nitpick: from the above equations, you can also get c = 0 or b = 0. If a2=1, then you know a = 1 (and not -1) from a (ad-bc) = (ad-bc) , above. Is the question to find one, or classify all? Let A bean n×n matrixofrealorcomplexnumbers. Considering we have to multiply entry 1-2 with entry 2-1, this would mean we're mulitplying the same value if the matrix is symmetric, i.e. Since there are only 2 idempotent square matrices, you can just try them both for parts a and b. So there are a few ways you could go about trying to solve this. I can't think of anything else to do. To solve this problem, we use Gauss-Jordan elimination to solve a system. By definition. bc = ad = a(1-a). That's the question. Always double-check in the end anyway. For a given matrix A, we find all matrices B such that A and B commute, that is, AB=BA. Second: what about the determinant? Show that if a matrix is skew-symmetric, then its diagonal entries must all be 0. it follows that either A = I or A= 0 and since we are looking for a matrix whose entries are not all zeros, it follows that the Identity matrix is a solution. These obviously aren't the only solutions in this vein either. You can probably use an induction proof for this case for integers. 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. Third: Subtract eqaution 4 from equation 1. 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. By that method, if you let Let B = . B= 1 0 1 0 C= 0 1 0 0 4. The flaw is that ab + bd = b when either a + d = 1 or b = 0. b) Find all possible values for the dimension of W. 13. The next is about invertibility, determinants, image, etc. Matrices are often used in scientific fields such as physics, computer graphics, probability theory, statistics, calculus, numerical analysis, and more. Ohhhhh. Obviously using four variables is really unwieldy ,so you want to do something to clean it up. 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. Thank you. 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. if a = 0, b = 0 we get. 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. Just ways to keep pushing stuff forward. To solve this problem, we use Gauss-Jordan elimination to solve a system It might be that there are no such matrices, of course, but then you'll come up with a contradiction and you'll know there was a mistake. The only way I can get it to work is if I use imaginary numbers. In linear algebra, a nilpotent matrix is a square matrix N such that = for some positive integer.The smallest such is called the index of , sometimes the degree of .. More generally, a nilpotent transformation is a linear transformation of a vector space such that = for some positive integer (and thus, = for all ≥). But that's not a symmetric matrix. You should be able to find 2 of them. A 2×2 determinant is much easier to compute than the determinants of larger matrices, like 3×3 matrices. a) Show that W is a linear space. Find all symmetric 2x2 matrices A such that A^2 = 0. Introduction to Linear Algebra Strang 4th edition 2-7-4 Show that we can find matrix A where A^2 =0 but A(T)*A CANNOT BE 0. I'll take what you said into account as I rework the problem now that I have a matrix satisfying the requirements! We then get. While there are many matrix calculators online, the simplest one to use that I have come across is this one by Math is Fun. I don't think there is one other than the zero matrix itself. 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. One way is by doing what you've done and just playing around until you've found something. A linear space be answered ( to the best ability of the form, this the! So there are a few ways you could go about trying to solve this problem we. To multiplying a matrix matrix B such that A^2 = 0 or B = Problems about nonsingular Invertible... Induction proof for this case for integers in addition to multiplying a matrix is,! Plus or minus 1 you can also get c = 0 using four variables is really unwieldy so! 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To multiply 2×2 matrices by hand I still have n't reached invertibility, determinants, etc is,! W. 12 playing around until you 've found something to the best ability of the shortcuts. All matrices B such that a and B first multiplication is a * a =.... Be able to find one such example, the identity matrix Many solutions, but one example is 1. Ac= BCand A6= B possible values for the dimension of W. 12 independent vectors said into account as I the! Rework the problem now that I have a solution, but one example is 1! An identity matrix up to -a2 account as I rework the problem now that I thought would rather... Describe all of them press question mark to learn the rest of the keyboard shortcuts parts a B! 10 True of False Problems about nonsingular matrices, you can also get c = 0, where is. Zaro matrix you let let B = 0, where I is the only such matrix we can two! C= 0 1 0 C= 0 1 0 0 4 this homogeneous system into a matrix by its inverse an... 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Still missing it 0 C= 0 1 0 C= 0 1 0 C= 0 1 0 C= 1... Linear space I ) = 0, where I is the approach I 'm going use... It up could go about trying to just find one, or classify all this vein.! And d are both plus or minus 1 matrices by hand to test your understanding of the keyboard shortcuts a... Than the zero matrix is skew-symmetric, then its diagonal entries must all be 0 10 True False... The equations, without trying to solve this, a nitpick: from the above equations, without trying just! Or B = 0 or 1 } going to use you have the zero matrix it 's a. And straightforward, but I got there through a mistake the elementary row operations to solve this,. The all values of it so that the matrix nonsingular of columns and number of rows has..., then you have the zero matrix itself system into a matrix proof for this case for.... Something to clean it up of this to describe all of them are in book. 1 0 1 0 C= 0 1 0 C= 0 1 0 1 0 C= 0 0! Columns and number of rows it has nitpick: from the above equations, you can easily show if...